Consider two circles C1 : x² + y² = 25 and C2 : (x - α)² + y² = 16, where α ∈ (5, 9). Let the angle

Q: Wait, why do we use the Law of Cosines on the triangle formed by the centers and the intersection point? Isn't there a simpler way?

That is a fantastic question! While you could potentially use the radical axis equation to find the intersection points, that approach often leads to messy coordinate geometry. By focusing on the triangle $\triangle C_1PC_2$, we leverage the inherent symmetry of the circles. The Law of Cosines is the most direct bridge between the side lengths (which depend on $\alpha$) and the given angle $\theta$. It turns a complex coordinate problem into a clean, elegant triangle problem.

Q: Why is the height of the triangle equal to $\frac{\beta}{2}$? I'm having trouble visualizing that.

Imagine the line segment connecting the two centers, $C_1$ and $C_2$. This line is the axis of symmetry for the two circles. Because the circles are symmetric about this line, the common chord must be perpendicular to this line and bisected by it. Since the common chord has total length $\beta$, the segment from the center-line to the intersection point $P$ must be exactly half of that, which is $\frac{\beta}{2}$. This is a classic geometric property that simplifies your calculations significantly!

Q: Why do we reject other possible values for $\alpha$? The problem says $\alpha \in (5, 9)$, but what if $\alpha$ could be something else?

The constraint $\alpha \in (5, 9)$ is not just a random range; it is a physical necessity for the circles to intersect at two distinct points. If $\alpha$ were too small (less than the difference of the radii) or too large (greater than the sum of the radii), the circles wouldn't intersect at all! When we solve $\alpha^2 = 36$, we get $\alpha = 6$ or $\alpha = -6$. Since $\alpha$ represents a distance and must fall within the given interval, $\alpha = 6$ is the only physically valid solution.

Question

Consider two circles $C_1 : x^2 + y^2 = 25$ and $C_2 : (x - \alpha)^2 + y^2 = 16$, where $\alpha \in (5, 9)$. Let the angle between the two radii (one to each circle) drawn from one of the intersection points of $C_1$ and $C_2$ be $\sin^{-1}(\frac{\sqrt{63}}{8})$. If the length of common chord of $C_1$ and $C_2$ is $\beta$, then the value of $(\alpha\beta)^2$ equals

Sigma Percentile · Accepted Answer

### The Geometry of Intersection: A Dance of Two Circles

Welcome, fellow explorer of the mathematical universe! Today, we are going to unravel a problem that might look like a standard coordinate geometry exercise, but beneath the surface, it is a beautiful dance of triangles and symmetry.

We are dealing with two circles, $C_1$ and $C_2$. $C_1$ is anchored at the origin with a radius of $5$, while $C_2$ is a wanderer, its center shifted along the x-axis by a distance $\alpha$.

Our goal is to find the value of $(\alpha\beta)^2$, where $\beta$ is the length of their common chord. Let us embark on this journey together.

### Phase 1: The Hidden Triangle

First, let us ground ourselves. We have $C_1: x^2 + y^2 = 25$ and $C_2: (x - \alpha)^2 + y^2 = 16$. The centers are $C_1(0, 0)$ and $C_2(\alpha, 0)$, with radii $r_1 = 5$ and $r_2 = 4$.

Now, imagine picking one of the intersection points, $P$. If we draw the radii from the centers to $P$, we create a triangle $	riangle C_1PC_2$.

The sides of this triangle are $r_1 = 5$, $r_2 = 4$, and the distance between the centers, which is $\alpha$. This triangle is the key to everything!

### Phase 2: The Law of Cosines

The problem gifts us with the angle between the radii at point $P$, which is $	heta = \sin^{-1}(\frac{\sqrt{63}}{8})$. To use this in our triangle, we need $\cos 	heta$.

Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$, we find $\cos^2 	heta = 1 - \frac{63}{64} = \frac{1}{64}$, so $\cos 	heta = \frac{1}{8}$.

Now, we apply the Law of Cosines to $	riangle C_1PC_2$:

$$ (C_1C_2)^2 = (C_1P)^2 + (C_2P)^2 - 2(C_1P)(C_2P) \cos 	heta $$

Substituting our values:

$$ \alpha^2 = 5^2 + 4^2 - 2(5)(4) \left( \frac{1}{8} ight) $$

This simplifies beautifully:

$$ \alpha^2 = 25 + 16 - 40 \left( \frac{1}{8} ight) = 41 - 5 = 36 $$

Thus, $\alpha = 6$. We have successfully tamed the first variable!

### Phase 3: The Common Chord and the Area Method

Now, we turn our attention to the common chord of length $\beta$. A fundamental property of intersecting circles is that the line joining their centers perpendicularly bisects the common chord.

This means the altitude of our triangle $	riangle C_1PC_2$ from vertex $P$ to the base $C_1C_2$ is exactly $\frac{\beta}{2}$. We can calculate the area of this triangle in two ways.

First, using the sine formula:

$$ 	ext{Area} = \frac{1}{2} r_1 r_2 \sin 	heta = \frac{1}{2} (5)(4) \left( \frac{\sqrt{63}}{8} ight) = \frac{5\sqrt{63}}{4} $$

Second, using the base-height formula:

$$ 	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \alpha 	imes \frac{\beta}{2} = \frac{1}{2} 	imes 6 	imes \frac{\beta}{2} = \frac{3\beta}{2} $$

Equating these two expressions:

$$ \frac{3\beta}{2} = \frac{5\sqrt{63}}{4} \Rightarrow \beta = \frac{5\sqrt{63}}{6} $$

### The Grand Finale

We are almost there! The problem asks for $(\alpha\beta)^2$. We have $\alpha = 6$ and $\beta = \frac{5\sqrt{63}}{6}$.

Multiplying them:

$$ \alpha\beta = 6 	imes \frac{5\sqrt{63}}{6} = 5\sqrt{63} $$

Finally, squaring this result:

$$ (\alpha\beta)^2 = (5\sqrt{63})^2 = 25 	imes 63 = 1575 $$

And there we have it! Through the power of geometry and a little bit of trigonometry, we have arrived at the solution. The final answer is **1575**.

Enter Numerical Value:

Visualized Solution

Visualizing the Circles

Identifying Centers and Radii

Intersection and Radii

The Angle Between Radii

Converting $sin θ$ to $cos θ$

Applying the Law of Cosines

Substituting Values

Solving for $α$

The Common Chord Geometry

Area of Triangle (Method 1)

Area of Triangle (Method 2)

Equating Areas to Find $β$

Final Computation

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Intersection

Phase 1

Phase 2

Phase 3

The Grand Finale

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