Let C1 be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let C2 be

Q: Why is the center of $C_1$ at $(-3, -3)$ and not $(3, 3)$?

Great catch! Always check your quadrants. The third quadrant is the region where both $x$ and $y$ are negative. Since the circle touches both axes, the distance from the center to the $x$-axis is $|y|$ and to the $y$-axis is $|x|$. Both must equal the radius, 3. Thus, $(-3, -3)$ is the only coordinate pair that satisfies the condition of being in the third quadrant.

Q: Do I really need the section formula? Can't I just find the equation of the circles and solve them simultaneously?

Technically, you could, but that is a classic 'JEE trap'! Solving the intersection of two circle equations involves quadratic systems that are prone to messy calculation errors. The section formula is the elegant, 'JEE-smart' way to find the point of contact. It leverages the geometry of the problem rather than brute-forcing the algebra.

Q: Wait, why does the point of contact divide the line segment in the ratio $r_1 : r_2$?

Think about the geometry of the tangent. The normal to the tangent at the point of contact passes through both centers. The distance from $O_1$ to the point of contact $P$ is simply the radius $r_1$, and the distance from $P$ to $O_2$ is the radius $r_2$. Therefore, $P$ naturally partitions the segment $O_1O_2$ in the ratio of the radii. It is a beautiful property of tangent circles!

Question

Let $C_1$ be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let $C_2$ be the circle with centre $(1, 3)$ that touches $C_1$ externally at the point $(\alpha, \beta)$. If $(\beta - \alpha)^2 = \frac{m}{n}, 	ext{gcd}(m, n) = 1$, then $m + n$ is equal to :

Options:

(A) 9

(B) 13

(C) 22

(D) 31

Sigma Percentile · Accepted Answer

### The Geometry of Connection

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are exploring the elegant dance of two circles touching in the vastness of the Cartesian plane.

When you look at a problem like this, don't just see equations. See the shapes. See the symmetry. Let us embark on this journey together.

### Phase 1: The First Circle

We begin with $C_1$. It is nestled in the third quadrant, a place where both $x$ and $y$ are negative. It has a radius of $r_1 = 3$.

Because it kisses both the $x$-axis and the $y$-axis, its center must be exactly 3 units away from both. In the third quadrant, this forces the center $O_1$ to be at $(-3, -3)$.

This is our anchor point. Without this, we are lost, but with it, we have a solid foundation.

### Phase 2: The Second Circle and the Bridge

Now, we introduce $C_2$. We know its center $O_2$ is at $(1, 3)$. We don't know its radius $r_2$ yet, but we know it touches $C_1$ externally.

When two circles touch externally, the distance between their centers, $d$, is the sum of their radii: $d = r_1 + r_2$. Let us calculate this distance $d$ using the distance formula:

$$d = \sqrt{(1 - (-3))^2 + (3 - (-3))^2}$$

This simplifies to:

$$d = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} = 2\sqrt{13}$$

Since $d = r_1 + r_2$, we have $2\sqrt{13} = 3 + r_2$, which gives us $r_2 = 2\sqrt{13} - 3$.

### Phase 3: The Point of Contact

The point of contact $P(\alpha, \beta)$ lies on the line segment connecting $O_1$ and $O_2$. Because the circles touch externally, this point $P$ divides the segment $O_1O_2$ internally in the ratio of their radii, $r_1 : r_2$.

This is the section formula in action. We are dividing the segment in the ratio $3 : (2\sqrt{13} - 3)$.

### Phase 4: The Algebraic Symphony

Let us calculate $\alpha$ and $\beta$. Using the section formula, $\alpha = \frac{r_1 x_2 + r_2 x_1}{r_1 + r_2}$. Substituting our values:

$$\alpha = \frac{3(1) + (2\sqrt{13}-3)(-3)}{2\sqrt{13}}$$

Expanding this, we get:

$$\alpha = \frac{3 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{12 - 6\sqrt{13}}{2\sqrt{13}} = \frac{6 - 3\sqrt{13}}{\sqrt{13}}$$

Similarly, for $\beta$:

$$\beta = \frac{3(3) + (2\sqrt{13}-3)(-3)}{2\sqrt{13}} = \frac{9 - 6\sqrt{13} + 9}{2\sqrt{13}} = \frac{18 - 6\sqrt{13}}{2\sqrt{13}} = \frac{9 - 3\sqrt{13}}{\sqrt{13}}$$

### Phase 5: The Final Resolution

Now, we look for $\beta - \alpha$. Notice the beauty of the subtraction:

$$\beta - \alpha = \frac{9 - 3\sqrt{13} - (6 - 3\sqrt{13})}{\sqrt{13}}$$

The term $-3\sqrt{13}$ cancels out perfectly! We are left with:

$$\beta - \alpha = \frac{9 - 6}{\sqrt{13}} = \frac{3}{\sqrt{13}}$$

Finally, the problem asks for the square of this difference:

$$(\beta - \alpha)^2 = \left(\frac{3}{\sqrt{13}}ight)^2 = \frac{9}{13}$$

Here, $m = 9$ and $n = 13$. Since $	ext{gcd}(9, 13) = 1$, our values are confirmed. The final answer, **$m + n$**, is **$9 + 13 = 22$**.

Select Answer:

Visualized Solution

Identifying Circle $C_{1}$

Properties of Circle $C_{2}$

Distance Between Centers

Calculating Distance $d$

External Touch Condition

Section Formula Setup

Calculating $α$

Calculating $β$

Finding $β - α$

The Final Calculation

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of Connection

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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