Let the circles C1 : (x - α)² + (y - β)² = r1² and C2 : (x - 8)² + (y - 15/2)² = r2² touch each othe

Q: Wait, why does the point of contact divide the line segment joining the centers? Is that always true?

Yes, absolutely! When two circles touch each other externally, the point of contact $P$ must lie on the line segment connecting the two centers. This is a fundamental geometric property. If they were touching internally, the point of contact would still be collinear with the centers, but it would lie on the extension of the line segment, not between them. Here, the problem explicitly states it's an internal division, confirming our geometric intuition.

Q: Why did we use the ratio $2:1$ for the section formula? How do I know which center corresponds to which part of the ratio?

Great question! The problem states that the point $P$ divides the line segment joining the centers of $C_1$ and $C_2$ in the ratio $2:1$. In the section formula $P = \frac{m \cdot c_2 + n \cdot c_1}{m+n}$, the ratio $m:n$ corresponds to the segments $c_1 P : P c_2$. Since the problem says $P$ divides the segment $c_1 c_2$ in ratio $2:1$, it means the distance from $c_1$ to $P$ is twice the distance from $P$ to $c_2$. Thus, $m=2$ and $n=1$.

Q: Can I use the property $r_1 + r_2 = \text{distance}(c_1, c_2)$ instead of the distance formula?

You certainly can! That is a valid property for externally touching circles. However, since you need the coordinates of the center $c_1(\alpha, \beta)$ to solve the problem anyway, using the distance formula from $P$ to $c_1$ to find $r_1$ is actually more direct. Both methods will lead you to the same result, so choose the one that feels most intuitive to you!

Question

Let the circles $C_1 : (x - \alpha)^2 + (y - \beta)^2 = r_1^2$ and $C_2 : (x - 8)^2 + (y - \frac{15}{2})^2 = r_2^2$ touch each other externally at the point $(6, 6)$. If the point $(6, 6)$ divides the line segment joining the centres of the circles $C_1$ and $C_2$ internally in the ratio $2 : 1$, then $(\alpha + \beta) + 4 (r_1^2 + r_2^2)$ equals

Options:

(A) 125

(B) 130

(C) 110

(D) 145

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, flat plane. Before you, two circles are drawn, each with its own unique center and radius. They are not overlapping, nor are they drifting apart in the void.

Instead, they are locked in a perfect, delicate embrace, touching at exactly one point: $P(6, 6)$. This is the beauty of coordinate geometry—what seems like a simple visual can be translated into a precise algebraic language.

Our mission is to uncover the hidden parameters of these circles, $\alpha$, $\beta$, $r_1$, and $r_2$, and combine them into a final, elegant value.

### Decoding the Centers

First, let us look at the circle $C_2$. Its equation is given as $(x - 8)^2 + (y - \frac{15}{2})^2 = r_2^2$.

In the language of geometry, the standard form of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center. By simply glancing at the equation, we can extract the center of $C_2$ as $c_2(8, \frac{15}{2})$.

Now, what about $C_1$? Its center $c_1(\alpha, \beta)$ is shrouded in mystery. But we have a clue: the point of contact $P(6, 6)$ acts as a bridge, dividing the line segment $c_1 c_2$ in a specific ratio of $2:1$.

### The Power of the Section Formula

This is where our mathematical toolkit becomes essential. The section formula is our bridge. If a point $P$ divides the segment joining $c_1$ and $c_2$ in the ratio $m:n$, its coordinates are given by:

$$P = \frac{m \cdot c_2 + n \cdot c_1}{m+n}$$

Here, our ratio $m:n$ is $2:1$. Let us apply this to the x-coordinates:

$$6 = \frac{2(8) + 1(\alpha)}{2+1}$$

With a quick multiplication, we find $18 = 16 + \alpha$, which leads us to $\alpha = 2$. Now, for the y-coordinates:

$$6 = \frac{2(\frac{15}{2}) + 1(\beta)}{2+1}$$

This simplifies to $18 = 15 + \beta$, giving us $\beta = 3$. We have successfully unmasked the center of $C_1$ as $(2, 3)$.

### Calculating the Radii

With the centers $c_1(2, 3)$ and $c_2(8, \frac{15}{2})$ firmly in our grasp, finding the radii is a straightforward journey. The radius $r_1$ is simply the distance from the center $c_1$ to the point of contact $P(6, 6)$.

Using the distance formula:

$$r_1^2 = (6 - 2)^2 + (6 - 3)^2 = 4^2 + 3^2 = 16 + 9 = 25$$

Similarly, for $r_2^2$, we calculate the distance from $c_2$ to $P$:

$$r_2^2 = (8 - 6)^2 + (\frac{15}{2} - 6)^2 = 2^2 + (\frac{3}{2})^2 = 4 + \frac{9}{4} = \frac{25}{4}$$

### The Final Synthesis

We have all the pieces of our puzzle: $\alpha = 2$, $\beta = 3$, $r_1^2 = 25$, and $r_2^2 = \frac{25}{4}$. Our target expression is $(\alpha + \beta) + 4(r_1^2 + r_2^2)$.

Substituting our values:

$$(2 + 3) + 4\left(25 + \frac{25}{4}\right)$$

This simplifies to $5 + 100 + 25$, which equals **130**. It is a moment of pure satisfaction when the variables align and the arithmetic resolves into a clean, final integer.

Select Answer:

Visualized Solution

Visualizing the Problem

Center of $C_{2}$

The Unknown Circle $C_{1}$

Section Formula

Equation for $α$

Solving for $α$

Equation for $β$

Solving for $β$

Calculating $r_{1}^{2}$

Calculating $r_{2}^{2}$

Evaluating the Expression

Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Decoding the Centers

The Power of the Section Formula

Calculating the Radii

The Final Synthesis

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Select Answer:

Visualized Solution

Visualizing the Problem

Center of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

The Unknown Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​