<think>
Okay, so I need to figure out how to derive a formula for the area of possible center points of all circles where points A and B lie on the circumference and point C is inside the area of the circle. Let me start by understanding the problem step by step.

First, let's denote the coordinates of points A, B, and C as follows:
- A = (x₁, y₁)
- B = (x₂, y₂)
- C = (x₃, y₃)

I need to find all possible centers (h, k) such that both A and B lie on the circumference of the circle with center (h, k), and point C lies inside this circle. Then, I have to determine the area covered by all such centers (h, k).

Let me recall some concepts from geometry. For a circle with center (h, k), the condition that points A and B lie on its circumference can be written using the distance formula. Specifically, for point A:

√[(x₁ - h)² + (y₁ - k)²] = r

Similarly, for point B:

√[(x₂ - h)² + (y₂ - k)²] = r

Here, r is the radius of the circle. Since both points lie on the circle, these equations must hold true simultaneously.

Since both equalities are equal to r, I can set them equal to each other:

√[(x₁ - h)² + (y₁ - k)²] = √[(x₂ - h)² + (y₂ - k)²]

Squaring both sides to eliminate the square roots gives:

(x₁ - h)² + (y₁ - k)² = (x₂ - h)² + (y₂ - k)²

Expanding both sides:

(x₁² - 2x₁h + h²) + (y₁² - 2y₁k + k²) = (x₂² - 2x₂h + h²) + (y₂² - 2y₂k + k²)

Simplify by subtracting h² and k² from both sides:

x₁² - 2x₁h + y₁² - 2y₁k = x₂² - 2x₂h + y₂² - 2y₂k

Now, let's bring like terms to one side:

(x₁² - x₂²) + (y₁² - y₂²) - 2h(x₁ - x₂) - 2k(y₁ - y₂) = 0

Let me factor the differences of squares:

[(x₁ - x₂)(x₁ + x₂)] + [(y₁ - y₂)(y₁ + y₂)] - 2h(x₁ - x₂) - 2k(y₁ - y₂) = 0

Hmm, perhaps I can factor out (x₁ - x₂) and (y₁ - y₂):

(x₁ - x₂)(x₁ + x₂ - 2h) + (y₁ - y₂)(y₁ + y₂ - 2k) = 0

Wait, that seems a bit messy. Let me rearrange the equation:

Let’s denote D_x = x₁ - x₂ and D_y = y₁ - y₂.

Then the equation becomes:

D_x(x₁ + x₂ - 2h) + D_y(y₁ + y₂ - 2k) = 0

So,

D_x (x₁ + x₂) - 2 D_x h + D_y (y₁ + y₂) - 2 D_y k = 0

Let me group the terms with h and k:

-2 D_x h - 2 D_y k + [D_x (x₁ + x₂) + D_y (y₁ + y₂)] = 0

Rearranged, this is:

2 D_x h + 2 D_y k = D_x (x₁ + x₂) + D_y (y₁ + y₂)

Divide both sides by 2:

D_x h + D_y k = (1/2)(D_x (x₁ + x₂) + D_y (y₁ + y₂))

So, this is the equation that relates h and k for points A and B to lie on the circle.

Let me denote the right-hand side as a constant for simplicity:

Let’s say,

K = (1/2)(D_x (x₁ + x₂) + D_y (y₁ + y₂))

So, the equation becomes:

D_x h + D_y k = K

This is a linear equation in variables h and k. So, for each circle passing through A and B, its center (h, k) must lie on this line.

But we also have the condition that point C lies inside the circle. For C to be inside the circle, the distance from C to the center (h, k) must be less than the radius r.

Given that r is the distance from A to the center:

r = √[(x₁ - h)² + (y₁ - k)²]

Similarly, for point C:

√[(x₃ - h)² + (y₃ - k)²] < r

Squaring both sides:

(x₃ - h)² + (y₃ - k)² < (x₁ - h)² + (y₁ - k)²

Expanding both sides:

x₃² - 2x₃h + h² + y₃² - 2y₃k + k² < x₁² - 2x₁h + h² + y₁² - 2y₁k + k²

Simplify by subtracting h² and k² from both sides:

x₃² - 2x₃h + y₃² - 2y₃k < x₁² - 2x₁h + y₁² - 2y₁k

Bring all terms to the left-hand side:

(x₃² - x₁²) + (y₃² - y₁²) - 2h(x₃ - x₁) - 2k(y₃ - y₁) < 0

Again, let me factor differences of squares:

[(x₃ - x₁)(x₃ + x₁)] + [(y₃ - y₁)(y₃ + y₁)] - 2h(x₃ - x₁) - 2k(y₃ - y₁) < 0

Let’s denote E_x = x₃ - x₁ and E_y = y₃ - y₁.

Then,

E_x (x₃ + x₁) + E_y (y₃ + y₁) - 2h E_x - 2k E_y < 0

Grouping terms with h and k:

-2 E_x h - 2 E_y k + [E_x (x₃ + x₁) + E_y (y₃ + y₁)] < 0

Which can be rewritten as:

2 E_x h + 2 E_y k > E_x (x₃ + x₁) + E_y (y₃ + y₁)

Divide both sides by 2:

E_x h + E_y k > (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁))

Let me denote the right-hand side as another constant:

Let’s say,

M = (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁))

So, we have:

E_x h + E_y k > M

Now, recall from earlier that for points A and B to lie on the circle, we had:

D_x h + D_y k = K

Where,

D_x = x₁ - x₂

D_y = y₁ - y₂

K = (1/2)(D_x (x₁ + x₂) + D_y (y₁ + y₂))

So now, the centers (h, k) must satisfy two conditions:

1. D_x h + D_y k = K
2. E_x h + E_y k > M

Where,

E_x = x₃ - x₁

E_y = y₃ - y₁

M = (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁))

So, we have two inequalities:

Equation 1: D_x h + D_y k = K
Inequality 2: E_x h + E_y k > M

So, the set of centers is a line for condition 1 and the region above the line defined by condition 2.

Therefore, the area of all possible centers (h, k) is the area of the region on one side of the line D_x h + D_y k = K, where E_x h + E_y k > M.

But this seems a bit abstract. Let me try to visualize it.

The first equation represents a straight line in the (h, k) plane. The inequality represents another half-plane. So, the set of all centers is the intersection of these two conditions: the center lies on the line and above the other line.

Wait, but if we have both D_x h + D_y k = K and E_x h + E_y k > M, then it's not just a single line; rather, it's points lying on one line and in another half-plane. So, actually, the set of centers is either empty or a ray (a line segment or an infinite ray) depending on whether the two lines intersect.

Wait, but if we have two different linear equations:

1. D_x h + D_y k = K
2. E_x h + E_y k > M

Then, unless both equations are identical, which would only happen if D_x = E_x and D_y = E_y, the first equation is a line, and the second inequality represents all points above another line.

But since A, B, C are arbitrary points (unless they're colinear or something), these lines will be different.

Therefore, depending on whether point C lies inside or outside certain regions relative to AB, the set of centers could be bounded or unbounded.

Wait, but if we have two linear equations in h and k, unless they represent parallel lines, which would create a strip, otherwise, they intersect at a single point. So, perhaps the area is between these two lines?

Wait, no. Because condition 1: D_x h + D_y k = K defines a line.

Condition 2: E_x h + E_y k > M defines half-plane above another line.

So, their intersection would be either empty or a ray.

Wait, let me think again.

If I have:

Equation 1: Line L1: D_x h + D_y k = K

Inequality 2: Line L2: E_x h + E_y k = M'

Where M' is such that E_x h + E_y k > M. So, inequality is above the line L2.

So, if L1 and L2 are not parallel, they intersect at a point P.

If I consider the set of points (h, k) lying on L1 and above L2, it's either empty or a ray starting from P going upwards in the direction determined by the inequality.

Wait, but actually, no. Because for each circle passing through A and B, its center lies on L1. So, all such centers are along line L1.

But then, point C has to lie inside the circle, which imposes another condition: that the distance from C to (h, k) is less than the radius.

Which led us to the inequality E_x h + E_y k > M.

So, since the center lies on L1, we substitute h and k in terms of a parameter along L1 into the inequality and see where it holds.

Alternatively, perhaps parametrize the line L1.

Let me attempt that.

First, from equation 1:

D_x h + D_y k = K

Let me solve for one variable, say h, in terms of k:

h = (K - D_y k) / D_x

Assuming D_x ≠ 0; if D_x is zero and D_y is non-zero, we can solve similarly.

So, substituting this into inequality 2:

E_x h + E_y k > M

Plugging in h:

E_x * [(K - D_y k)/D_x] + E_y k > M

Multiply through by D_x (assuming D_x ≠ 0):

E_x (K - D_y k) + E_y D_x k > M D_x

Expanding:

E_x K - E_x D_y k + E_y D_x k > M D_x

Grouping terms with k:

(-E_x D_y + E_y D_x)k + E_x K > M D_x

So,

[ (E_y D_x - E_x D_y ) ] k + E_x K > M D_x

Let me denote the coefficient of k as C = E_y D_x - E_x D_y.

Then, we have:

C k + E_x K > M D_x

Solve for k:

k > (M D_x - E_x K)/C

Assuming C ≠ 0; if C is zero, then it's a different case.

So, assuming C ≠ 0,

k > [ (M D_x - E_x K ) ] / C

But M is defined as:

M = (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁))

Similarly, K was defined earlier.

So, let me plug in K and M into the inequality:

k > [ ( (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁)) * D_x - E_x K ) ] / C

But this is getting complicated. Let me see if I can express it differently.

Alternatively, perhaps think of both L1 and L2 as lines in the h-k plane.

Given that centers lie on L1 and satisfy the inequality relative to L2, their intersection would form a region. If L1 and L2 intersect at point P, then depending on which side of L2 we are (due to the inequality), the set of centers is either empty or a ray starting from P in some direction.

Therefore, if I can find whether point P satisfies E_x h + E_y k > M or not, that would determine if there's a possible center.

Wait, but maybe instead of getting bogged down with algebra, let me think geometrically.

Given points A and B, the set of circles passing through them have centers lying on the perpendicular bisector of segment AB. So, equation D_x h + D_y k = K is indeed the perpendicular bisector.

Now, point C must lie inside such a circle. Therefore, for each center (h, k) on the perpendicular bisector, we need to ensure that distance from (h, k) to C is less than radius r.

But since r is equal to the distance from (h, k) to A (or B), so the condition becomes:

Distance from (h, k) to C < Distance from (h, k) to A

Which can be written as:

√[(x₃ - h)² + (y₃ - k)²] < √[(x₁ - h)² + (y₁ - k)²]

Squaring both sides:

(x₃ - h)^2 + (y₃ - k)^2 < (x₁ - h)^2 + (y₁ - k)^2

Expanding:

x₃² - 2x₃h + h² + y₃² - 2y₃k + k² < x₁² - 2x₁h + h² + y₁² - 2y₁k + k²

Simplify by subtracting h² and k² from both sides:

x₃² - 2x₃h + y₃² - 2y₃k < x₁² - 2x₁h + y₁² - 2y₁k

Bring all terms to left-hand side:

(x₃² - x₁²) + (y₃² - y₁²) - 2h(x₃ - x₁) - 2k(y₃ - y₁) < 0

Factor differences of squares:

[(x₃ - x₁)(x₃ + x₁)] + [(y₃ - y₁)(y₃ + y₁)] - 2h(x₃ - x₁) - 2k(y₃ - y₁) < 0

So, this is similar to what I had earlier.

Let me denote E_x = x₃ - x₁ and E_y = y₃ - y₁.

Then:

E_x (x₃ + x₁) + E_y (y₃ + y₁) - 2h E_x - 2k E_y < 0

Which can be rewritten as:

-2 E_x h - 2 E_y k + [E_x (x₃ + x₁) + E_y (y₃ + y₁)] < 0

Divide by -2 (reversing the inequality):

E_x h + E_y k > (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁))

Which is what I had before.

So, we have two conditions:

1. D_x h + D_y k = K

2. E_x h + E_y k > M

Where,

D_x = x₁ - x₂

D_y = y₁ - y₂

K = (1/2)(D_x (x₁ + x₂) + D_y (y₁ + y₂))

E_x = x₃ - x₁

E_y = y₃ - y₁

M = (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁))

So, now I need to find the area of centers (h, k) that lie on L1 and satisfy E_x h + E_y k > M.

Since L1 is a straight line, and the inequality defines a half-plane, their intersection would be either empty or a ray.

Therefore, if point P, the intersection of L1 and L2, satisfies the inequality, then the region is a ray starting from P. Otherwise, it's empty.

But to find the area, which is zero for an empty set or infinite for a ray? But in our case, we are dealing with all possible centers that satisfy these conditions. So, depending on whether such centers exist, the area could be zero (if no solutions) or unbounded (if infinitely many). However, since A and B are fixed, and C is given, perhaps it's more likely to have a bounded region.

Wait, maybe I made an error in thinking that the set of centers is a ray. Let me reconsider.

Actually, for each circle passing through A and B, the center lies on L1 (the perpendicular bisector). For such circles to contain C inside them, the distance from C to the center must be less than the radius. But the radius is equal to the distance from A or B to the center.

So, it's equivalent to |C - center| < |A - center|. Which can be rewritten as:

|C - center|² < |A - center|²

Which simplifies to:

(x₃ - h)^2 + (y₃ - k)^2 < (x₁ - h)^2 + (y₁ - k)^2

As before.

Expanding:

x₃² - 2x₃h + y₃² - 2y₃k < x₁² - 2x₁h + y₁² - 2y₁k

Which simplifies to:

(x₃² - x₁²) + (y₃² - y₁²) - 2h(x₃ - x₁) - 2k(y₃ - y₁) < 0

So, it's a linear inequality in h and k. So, the set of centers is the intersection of L1 and the half-plane defined by this inequality.

Thus, if I have two lines L1: D_x h + D_y k = K and L2: E_x h + E_y k = M', then their intersection defines point P. The region where both conditions hold would either be empty or a single point, depending on whether C is inside the circle.

Wait, that can't be right because if I think about it, for some positions of C relative to AB, there might be an area of centers such that C lies inside all such circles.

But wait, actually, as you vary the center along L1, the radius varies as well. So, maybe when moving the center closer or farther from A and B, the condition on C can flip between being inside or outside.

Wait, perhaps it's better to think of the locus of centers (h, k) such that both A and B lie on their circle and C is inside. Since both A and B are fixed, the circles passing through them have their centers moving along L1.

So, as the center moves along L1, the radius changes. The point C will be inside the circle if the distance from C to the center is less than the radius.

But since the radius is equal to the distance from A (or B) to the center, that condition can be rewritten as |C - center| < |A - center|. Which is equivalent to the power of point C with respect to the circle centered at center passing through A and B. The power condition tells us that if |C - center|² - |A - center|² < 0, then C lies inside.

Wait, let me compute that:

|C - center|² - |A - center|² = (x₃ - h)^2 + (y₃ - k)^2 - [(x₁ - h)^2 + (y₁ - k)^2]

Expanding:

= x₃² - 2x₃h + y₃² - 2y₃k - x₁² + 2x₁h - y₁² + 2y₁k

Simplify:

(x₃² - x₁²) + (y₃² - y₁²) + (-2x₃h + 2x₁h) + (-2y₃k + 2y₁k)

= (x₃ - x₁)(x₃ + x₁) + (y₃ - y₁)(y₃ + y₁) + 2h(x₁ - x₃) + 2k(y₁ - y₃)

Which is:

E_x (x₃ + x₁) + E_y (y₃ + y₁) - 2h E_x - 2k E_y

Wait, which was the same as before.

So, for C to lie inside, this expression must be less than zero:

E_x (x₃ + x₁) + E_y (y₃ + y₁) - 2h E_x - 2k E_y < 0

Which can be written as:

- [2 E_x h + 2 E_y k] + [E_x (x₃ + x₁) + E_y (y₃ + y₁)] < 0

Or,

2 E_x h + 2 E_y k > E_x (x₃ + x₁) + E_y (y₃ + y₁)

Divide both sides by 2:

E_x h + E_y k > M'

Where M' = (1/2)(E_x (x₃ + x₁) + E_y (y₃ + y₁))

So, as before.

Now, since we have two linear equations:

1. D_x h + D_y k = K
2. E_x h + E_y k > M'

These can be thought of as lines in the h-k plane. The first is fixed, and the second is an inequality defining a half-plane.

So, to find the set of centers (h, k), we need to find all points on line L1 where E_x h + E_y k > M'.

Graphically, if I imagine line L1, which is the perpendicular bisector of AB, then the inequality E_x h + E_y k > M' defines a half-plane. The intersection of these two will be either empty or a ray.

Therefore, the set of centers is either empty or a ray in the h-k plane.

If it's a ray, the area covered by all such centers would technically be infinite because there are infinitely many points along a ray. However, since we need the "area" of possible centers, if it's unbounded, maybe the answer is that no finite area exists? But that seems counterintuitive because with A and B fixed, moving the center along L1 should create some locus.

Wait, perhaps I made an error in thinking about the problem. Maybe instead of considering all circles through A and B containing C, we need to find centers such that C is inside the circle, so it's not just a matter of varying the radius but ensuring that regardless of how you move the center along L1, C remains inside.

But no, actually, for each position of the center on L1, the circle will have a different radius. So, point C can be either inside or outside depending on where the center is. Therefore, there might be regions on L1 where moving the center causes C to enter or exit the circle.

Wait, but if you move the center along L1 towards A, the radius decreases, so C may or may not lie inside. Similarly, moving away from A increases the radius, potentially including C again.

Therefore, perhaps there are two intervals on L1 where C is inside. But since we’re dealing with an infinite line, maybe the area covered by all possible centers (h, k) is a bounded region?

Wait, no. Because as you move along L1, the center can go to infinity in either direction, but for each position, there's only one point (h, k). So, it's not exactly an area unless we consider some measure over those points.

Wait, perhaps I need to parameterize L1 and see how h and k relate as you move along it. Maybe express h and k in terms of a single variable and then integrate or find the envelope?

Alternatively, think about what happens when the center moves on L1. The condition for C being inside is a linear inequality, so the set of centers satisfying both conditions would be an intersection of two lines, resulting in either no solution or a single point if they are parallel.

But actually, as we saw before, it's either empty or a ray.

Wait, let me consider specific coordinates to test this.

Suppose A = (0, 0), B = (2, 0). So D_x = -2, D_y = 0. K = (-2)(0 + 2)/2 = -4.

So L1: -2 h + 0 k = -4 => -2h = -4 => h = 2.

Wait, that can't be right because if A is (0,0) and B is (2,0), the perpendicular bisector is x=1. Hmm, so maybe I made a mistake in calculating K.

Wait, no:

If D_x = x₁ - x₂ = 0 - 2 = -2

D_y = y₁ - y₂ = 0 - 0 = 0

K = (1/2)(D_x(x₁ + x₂) + D_y(y₁ + y₂)) = (1/2)(-2*(0 + 2) + 0*(0 + 0)) = (1/2)(-4) = -2.

So L1: -2h + 0k = -2 => h = 1. So, the perpendicular bisector is x=1, which makes sense for A(0,0) and B(2,0).

Now, let me choose C as (1,1). So E_x = 1 - 0 = 1, E_y = 1 - 0 = 1.

M' = (1/2)(1*(1 + 0) + 1*(1 + 0)) = (1/2)(1 + 1) = 1

So the inequality is:

E_x h + E_y k > M'

Which becomes:

1*h + 1*k > 1 => h + k > 1

So, on line L1: x=1, we have to find points where h + k > 1.

Since on L1, h = 1, so plug in h=1 into inequality:

1 + k > 1 => k > 0

Therefore, the set of centers is all points (1, k) where k > 0. So it's a ray starting at (1, 0) going upwards.

So, area-wise, this is an infinite vertical line segment starting from (1,0). But in terms of area, since it's one-dimensional, the "area" would be zero or unbounded? Hmm, but in two dimensions, it's a line, so its measure is zero. However, if we think about the set of all possible centers, each center corresponds to a unique circle through A and B containing C.

But wait, actually, for each h on L1 where k > 0, there is a corresponding circle with radius equal to distance from (h,k) to A or B. So, for our specific case:

Center at (1, k), so radius squared is (0 - 1)^2 + (0 - k)^2 = 1 + k².

Distance from C(1,1) to center: √[(1-1)^2 + (1 - k)^2] = |1 - k|

For C to be inside the circle:

|1 - k| < √(1 + k²)

Squaring both sides (since both are non-negative):

(1 - k)² < 1 + k²

Expanding:

1 - 2k + k² < 1 + k²

Subtract 1 and k² from both sides:

-2k < 0 => k > 0

Which is consistent with earlier result. So, indeed, for centers on L1 above (1,0), the condition holds.

So, in this specific case, all such centers lie along the line x=1, y > 0, which is a ray starting at (1,0). Therefore, the set of centers is not an area but a one-dimensional curve. So, the "area" covered by these centers would be zero because it's just a line segment.

But this contradicts my earlier thought that the answer might involve a non-zero finite area.

Wait, perhaps I need to think differently. Maybe in some cases, when C is inside a certain region relative to AB, the set of possible centers is bounded and forms an area.

Wait, let me try with another configuration. Suppose A = (0, 0), B = (2, 0), and C = (1, 2).

So E_x = 1 - 0 = 1, E_y = 2 - 0 = 2

M' = (1/2)(1*(1 + 0) + 2*(2 + 0)) = (1/2)(1 + 4) = (5)/2

So the inequality is:

h + 2k > 5/2

On L1, which is x=1.

So h=1, so substituting:

1 + 2k > 5/2 => 2k > 3/2 => k > 3/4

Therefore, centers are points (1, k) where k > 3/4. So again, it's a ray starting at (1, 3/4). So the area is still zero.

Wait, but maybe if C is on one side of AB, we might get a bounded region?

Let me try another configuration. Let’s take A = (0,0), B=(2,0), and C = (-1,1).

So E_x = -1 - 0 = -1, E_y = 1 - 0 = 1.

M' = (1/2)(-1*( -1 + 0 ) + 1*(1 + 0)) = (1/2)(-1*(-1) + 1*1) = (1/2)(1 + 1) = 1

So the inequality is:

(-1)h + (1)k > 1 => -h + k > 1

On L1: x=1.

Plug in h=1:

-1 + k > 1 => k > 2

Therefore, centers are points (1, k) where k > 2. So again, a ray starting at (1,2). The area is still zero.

Wait, perhaps I need to consider when the two lines L1 and L2 intersect in such a way that the region is bounded?

Suppose C is very close to A. Let’s take C=(0.1, 0.1).

So E_x = 0.1 - 0 = 0.1, E_y = 0.1 - 0 = 0.1

M' = (1/2)(0.1*(0.1 + 0) + 0.1*(0.1 + 0)) = (1/2)(0 + 0.01) = 0.005

So inequality is:

0.1 h + 0.1 k > 0.005

Multiply both sides by 10: h + k > 0.5

On L1, which is x=1.

Plug in h=1: 1 + k > 0.5 => k > -0.5

So centers are all points (1, k) where k > -0.5. So it's a ray starting at (1, -0.5). The area covered is again zero.

Wait, so in all these cases, the set of centers lie on a line L1 and satisfy another linear inequality, resulting in either an empty set or a ray. Therefore, the "area" is zero because it's a one-dimensional curve.

But that seems contradictory to the problem statement which asks for the area of possible center points. Maybe I misunderstood the question.

Wait, re-reading the problem: "derive a formula that for all possible values for A B and C gives the area of possible center points of all circles such that A and B are on the circumference and C is within the area of the circle."

So it's not about each individual circle, but all centers (h,k) such that for some circle through A and B containing C. So yes, in my examples above, the set of centers lies along L1 with a certain inequality constraint, leading to either empty or a ray.

But if we think about this in terms of possible positions (h,k), each center is a point on L1 satisfying E_x h + E_y k > M', which is also linear. So their intersection is either empty or a single point or a line. But in our cases, it's either empty or a ray.

Wait, but the problem says "area of possible center points", so if all centers lie along a line, then technically, their area is zero because a line has measure zero in two-dimensional space.

Alternatively, maybe I need to consider the set of circles through A and B containing C, which would correspond to regions on L1. But for each circle, there's exactly one center. So perhaps it's not about an area but rather the possible positions of centers, which lie along a line segment or a ray.

But in all my examples, it was either empty or a ray starting at some point. Therefore, maybe the answer is that the area is zero because all centers lie on a straight line.

Wait, but if you have two lines, their intersection is a point, so unless L1 and L2 coincide, which would make the set of centers unbounded along L1.

Alternatively, perhaps I made an error in interpreting E_x h + E_y k > M'. Maybe it's better to think geometrically about what locus this corresponds to.

Wait, let me recall that for a fixed point C, the condition |C - center| < |A - center| is equivalent to C lying inside the circle inversion of A with respect to the variable center. But I'm not sure if that helps.

Alternatively, using coordinate geometry, the inequality E_x h + E_y k > M' can be thought of as a line. So, L1 is the perpendicular bisector of AB, and this inequality defines another line. The intersection of these two lines will give specific centers where C lies on the circle. But if I want C to lie inside, then it's either empty or a ray.

Wait, actually, in all my examples, the set of centers was a ray because L1 is a straight line and the inequality defines another half-plane cutting through it. So their intersection is a ray.

But then, since each center corresponds to exactly one circle, the "area" covered by these centers would technically be zero as they lie on a line.

Therefore, perhaps the answer is that there's no area because all possible centers lie on a straight line, hence having measure zero in two-dimensional space. Therefore, the area is zero.

But let me see if I can think of another configuration where it might not be a ray or empty. Suppose C is very close to AB but above and below. Wait, suppose A=(0,0), B=(2,0), and C=(1,10). Then E_x=1-0=1, E_y=10-0=10.

M' = (1/2)(1*(1+0) + 10*(10 + 0)) = (1/2)(1 + 100) = 101/2 = 50.5

So the inequality is:

h + 10k > 50.5

On L1: h=1.

Thus, 1 + 10k > 50.5 => 10k > 49.5 => k > 4.95

Therefore, centers lie along (1, k) where k > ~4.95. So again, it's a ray starting at (1, ~4.95). The area covered is zero.

Wait, so regardless of the position of C relative to AB, the set of possible centers lies on L1 and satisfies another linear inequality, resulting in either empty or a ray. Therefore, the "area" is always zero because it's just a line segment (or ray) in two-dimensional space.

But then why does the problem ask for an area? Maybe I misinterpret something.

Wait, perhaps I need to consider all possible circles through A and B that contain C somewhere inside, not necessarily centered on L1. But no, by definition, all centers must lie on L1 because A and B are fixed points.

Alternatively, maybe it's a locus of centers such that the circle passes through A and B and contains C. So, for each center on L1, there is exactly one circle. Therefore, the set of centers is as we found: either empty or a ray or a line segment.

But in all cases, it seems to be either empty or a ray. Thus, the area covered by these centers would be zero because they lie along a straight line.

Therefore, perhaps the answer is that there is no finite area; the set of possible centers lies on a straight line, hence having measure zero in two-dimensional space.

But then again, maybe I need to compute something else. Wait, could it be that for all such circles containing C, their centers lie within some bounded region? But from my examples above, when moving along L1, the center can go far away, but as long as k is large enough, C remains inside.

Wait, no. If you move the center along L1 towards infinity, the radius also tends to infinity because it's equal to the distance from A (or B) to the center. So, even though C is fixed, moving the center far away would make the circle larger, so C will always lie inside as long as certain conditions are met.

Wait, but in my earlier examples, when I moved along L1 towards increasing k, it was a ray; but what if moving in the opposite direction?

Wait, no. For example, take A=(0,0), B=(2,0). If you move the center on L1 (x=1) to negative y-values, then the radius is sqrt(1 + k²). The distance from C to the center is |k - k_c|, assuming C is at (1, something).

Wait, for example, if C=(1, 2), moving the center down along L1: (1, k) with k < 0. Then the radius becomes sqrt(1 + k²). The distance from C to the center is |2 - k|. For C to be inside, we need |2 - k| < sqrt(1 + k²).

Which simplifies as before:

(2 - k)^2 < 1 + k²

4 - 4k + k² < 1 + k²

3 - 4k < 0 => k > 3/4.

So, even when moving the center down below (1,0), the condition requires k > 3/4 to keep C inside. Therefore, centers can't be too far in either direction; otherwise, C might exit the circle.

Wait, but in my previous calculation where I had A=(0,0), B=(2,0), and C=(1,2), moving the center down below (1,3/4) would cause C to be outside. So actually, for centers on L1, there's a lower bound on k.

Wait, so in that case, if we have L1: x=1, and inequality k > 3/4, then the set of centers is from (1, 3/4) upwards, which is a ray starting at (1, 3/4). So again, it's an infinite line segment.

But in another configuration where C is closer to AB, say C=(1,0.5), let's see:

E_x = 1 - 0 = 1

E_y = 0.5 - 0 = 0.5

M' = (1/2)(1*(1 + 0) + 0.5*(0.5 + 0)) = (1/2)(1 + 0.25) = 1.25 / 2 = 0.625.

So inequality is h + 0.5k > 0.625

On L1, h=1:

1 + 0.5k > 0.625 => 0.5k > -0.375 => k > -0.75

Therefore, centers are all points (1, k) where k > -0.75.

So it's a ray starting at (1, -0.75). So again, the set of possible centers is a line segment on L1 with one endpoint at (1, -0.75).

Wait, so in this case, if C is below AB, say C=(1,-1):

E_x = 1 - 0 = 1

E_y = -1 - 0 = -1

M' = (1/2)(1*(1 + 0) + (-1)*(-1 + 0)) = (1/2)(1 + 1) = 1.

So inequality is h - k > 1

On L1, h=1:

1 - k > 1 => -k > 0 => k < 0

Therefore, centers are all points (1, k) where k < 0. So it's a ray starting at (1, 0) going downwards.

But in this case, if you move the center below (1,0), will C=(1,-1) always lie inside?

Wait, let me compute:

Center is (1,k). Radius squared is 1 + k².

Distance from C to center: | -1 - k | = |k + 1|

For C to be inside, we need |k + 1| < sqrt(1 + k²)

Square both sides:

(k + 1)^2 < 1 + k²

Expanding:

k² + 2k + 1 < 1 + k²

Cancel terms: 2k < 0 => k < 0.

So, indeed, for all k < 0, C lies inside. Therefore, the set of centers is a ray starting at (1,0) going downward.

But again, it's an infinite line segment; thus, zero area.

Wait, so in all these cases, whether C is above or below AB, the set of possible centers is either empty, a single point, or a ray on L1. Hence, their "area" is always zero because they lie along a straight line.

But then why does the problem ask for an area? Maybe I need to think differently.

Wait, perhaps it's about the set of all circles through A and B containing C somewhere inside. But each circle corresponds to exactly one center on L1. So, the number of such centers is either zero or a ray (infinite points). Therefore, the "area" covered by these centers would indeed be zero.

Alternatively, perhaps I need to parameterize the movement of the center and integrate over some region? But in all my examples, it's just a line segment with no area.

Wait, maybe if C is inside a certain region relative to AB, then the locus of possible centers is bounded. Let me think about that.

Suppose A=(0,0), B=(2,0). If I take C somewhere above AB, say (1, d) where d > 0. Then E_x=1-0=1, E_y=d - 0 = d.

M' = (1/2)(1*(1 + 0) + d*(d + 0)) = (1/2)(1 + d²)

Inequality: h + d k > (1 + d²)/2

On L1, h=1:

1 + d k > (1 + d²)/2 => multiply both sides by 2: 2 + 2d k > 1 + d² => 2d k > -1 + d² => k > (-1 + d²)/(2d)

So the lower bound on k is k_min = (-1 + d²)/(2d). Let's see what this value is:

k_min = (d² - 1)/(2d)

If d > 1, then numerator positive: so k_min positive.

If 0 < d < 1, numerator negative: so k_min negative.

So depending on the position of C relative to AB, the lower bound on k varies.

But in any case, regardless of where C is (as long as it's not on L1), we can have a ray or an interval on L1. But since L1 is infinite in both directions, but our inequality cuts it at some point, making it either empty or a ray.

Therefore, the set of centers is always a one-dimensional object, meaning its area is zero.

Thus, my conclusion is that regardless of the positions of A, B, and C (as long as they're not colinear in a certain way), the possible centers lie on a straight line, making their "area" zero.

But this contradicts the problem statement which implies that such an area exists. Maybe I need to think about another approach.

Wait, perhaps it's not just about circles through A and B containing C, but circles where both A and B are on the circumference and C is inside. So for each circle passing through A and B with center on L1, if C is inside, then (h,k) must lie in some region.

But as we saw earlier, the condition E_x h + E_y k > M' cuts L1 into a ray or empty set. Therefore, the locus of possible centers is either empty or a ray, which again has measure zero in 2D space.

Alternatively, maybe if I consider all such circles, their centers form an ellipse or something else? Wait, no. Because for two points A and B, the set of centers of circles passing through them lie on L1, which is a line. So, the locus can't be anything else unless you have more constraints.

Wait, perhaps it's better to think in terms of inversion geometry.

But honestly, after going through multiple examples and calculations, I keep arriving at the conclusion that the set of possible centers lies along a straight line (either empty or a ray), hence their area is zero. Therefore, maybe the answer is simply zero.

Alternatively, if the problem expects an expression in terms of A, B, C, perhaps it's a line segment on L1 between certain points, but since we can't have a non-zero area from a one-dimensional set, perhaps the answer is that no such finite area exists, or that the area is zero.

But I'm not entirely sure. Maybe I need to think about other approaches.

Wait, another way: for all circles through A and B containing C, let's find the locus of their centers.

Given two fixed points A and B, any circle passing through them has its center on L1, the perpendicular bisector of AB.

For such a circle to contain C, we need |C - O| < |A - O|, where O is the center.

This condition can be rewritten as:

|C - O|² < |A - O|²

Which simplifies to:

O being closer to C than A.

In coordinate terms, if you fix A and B, L1 is known. For each point O on L1, check if |C - O| < |A - O|. The set of such points O forms a region on L1, which, as we saw earlier, is either empty or a ray.

Therefore, the locus of centers is indeed along L1 with certain constraints, meaning it's one-dimensional. Hence, its area is zero.

So, putting all this together, I think the answer is that the area of possible center points is zero because they lie on a straight line in the plane, which has measure zero in two dimensions.

But just to be thorough, let me see if there exists any configuration where the set of centers isn't a ray or empty. Suppose A, B, and C are colinear with AB being the segment from A=(0,0) to B=(2,0), and C is somewhere else on the line AB beyond B, say C=(3,0).

Then E_x = 3 - 0 = 3, E_y=0-0=0.

M' = (1/2)(3*(3 + 0) + 0*(0 + 0)) = (1/2)(9 + 0) = 4.5

Inequality: 3h + 0k > 4.5 => 3h > 4.5 => h > 1.5.

On L1, which is x=1.

Plug in h=1:

1 * 3 + 0 * k = 3 > 4.5? No. So no solution. Therefore, centers must lie where?

Wait, on L1: x=1, so plug into inequality:

3*1 + 0*k > 4.5 => 3 > 4.5? No, which is false.

Thus, no such centers exist because C is too far along AB beyond B, making it impossible for any circle through A and B to contain C without the center being closer than possible.

Therefore, in this case, the set of centers is empty.

But if I place C close enough:

Take C=(3, ε), very small ε>0.

E_x=3-0=3

E_y=ε - 0=ε

M' = (1/2)(3*(3 + 0) + ε*(ε + 0)) = (9/2) + something negligible ≈ 4.5.

Inequality: 3h + εk > 4.5

On L1, h=1:

3*1 + εk > 4.5 => 3 + εk > 4.5 => εk > 1.5.

But since ε is very small, say ε=0.1, then k must be greater than 15, which is impossible because the center on L1 would have to be at (1,15), but moving that far away doesn't contain C=(3,0.1).

Wait, let me compute:

Center O=(1,15). Radius squared: 1 + 225=226.

Distance from C to O: sqrt((3-1)^2 + (0.1 -15)^2)=sqrt(4 + 220.9)=sqrt(224.9)≈14.99

Radius is sqrt(226)≈15.03, so distance from C to O is ≈14.99 < 15.03.

Thus, C=(3,0.1) lies inside the circle centered at (1,15). So even though C is far along AB, a sufficiently large center can still contain it.

Wait, but according to our inequality earlier:

E_x h + E_y k > M'

Which for C=(3,0.1):

3*1 + 0.1*k > 4.5

So 3 + 0.1k > 4.5 => 0.1k > 1.5 => k>15.

So yes, centers need to be above (1,15) to contain C=(3,0.1). Therefore, the set of centers is a ray starting at (1,15).

But again, this is an infinite line segment; hence, area zero.

Therefore, in all cases I can think of, the locus of possible centers lies on a straight line, which has zero area. Therefore, my conclusion is that the area of possible center points is zero.

**Final Answer**
The area of possible center points is \boxed{0}.
</think>

Given points \(A\), \(B\), and \(C\) in a plane, we need to derive a formula for the area of possible center points of all circles such that \(A\) and \(B\) are on the circumference and \(C\) is within the area of the circle.

1. **Points Definition**:
   - Let \(A = (x_A, y_A)\), \(B = (x_B, y_B)\), and \(C = (x_C, y_C)\).

2. **Perpendicular Bisector**:
   - The set of all centers of circles passing through \(A\) and \(B\) lies on the perpendicular bisector \(L_1\) of segment \(AB\). This line is given by \(L_1: x = \frac{x_A + x_B}{2}\).

3. **Condition for Containment**:
   - For a circle with center \((h, k)\) to contain point \(C\), the distance from \(C\) to the center must be less than the radius of the circle.
   - This translates to the inequality: \(|C - O|^2 < |A - O|^2\).

4. **Simplifying the Inequality**:
   - Substituting \(O = (h, k)\), we get:
     \[
     (x_C - h)^2 + (y_C - k)^2 < (x_A - h)^2 + (y_A - k)^2
     \]
   - Expanding and simplifying this inequality results in a linear constraint on \(h\) and \(k\).

5. **Intersection of Lines**:
   - The perpendicular bisector \(L_1\) intersects with the line derived from the inequality, resulting in either an empty set or a ray on \(L_1\).

6. **Conclusion**:
   - The set of possible centers lies along a straight line (either empty or a ray), which has measure zero in two-dimensional space.
   - Therefore, the area covered by these centers is zero.

The final answer is:
\[
\boxed{0}
\]