JEE Main: Circles Touching Axes & Each Other — Smart Geometry Method 💡

 

❓ Question

FOR:

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

$$(\beta - \alpha)^2 = \frac{m}{n}, \quad \gcd(m,n)=1,$$

then the value of

$$m+n$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This problem uses pure coordinate geometry logic:

👉 A circle touching both axes has its centre fixed by symmetry.
👉 When two circles touch externally, the point of contact lies on the line joining their centres.
👉 We find that point using section formula, then compute $(\beta-\alpha)^2$.

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🔹 Step 1 — Equation and centre of $C_1$

Circle $C_1$:

• Lies in third quadrant

• Touches x-axis and y-axis

• Radius = 3

Hence, its centre must be:

$$(-3,\,-3)$$

(Only this point keeps the circle in the third quadrant and tangent to both axes.)

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🔹 Step 2 — Centre and radius relation with $C_2$

Given:

• Centre of $C_2 = (1,\,3)$
  

• $C_2$ touches $C_1$ externally

Distance between centres:

$$\sqrt{(1+3{)}^2+(3+3{)}^2}=\sqrt{4^2+6^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{\mathrm{13}}$$

Let radius of $C_2 = r$.

External touching condition:

$$3 + r = 2\sqrt{13} \Rightarrow r = 2\sqrt{13} - 3$$

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🔹 Step 3 — Coordinates of point of contact $(\alpha,\beta)$

For external contact, the point divides the line joining the centres internally in the ratio of radii:

$$C_1C_2 = 3 : r$$

Using section formula between:

$$(-3,-3)\ \text{and}\ (1,3)$$
$$\alpha = \frac{3(1) + r(-3)}{3 + r}, \quad \beta = \frac{3(3) + r(-3)}{3 + r}$$

Substitute $r = 2\sqrt{13}-3$​:

$$\alpha = \frac{3 - 3(2\sqrt{13}-3)}{2\sqrt{13}} = \frac{12 - 6\sqrt{13}}{2\sqrt{13}}$$ $$\beta = \frac{9 - 3(2\sqrt{13}-3)}{2\sqrt{13}} = \frac{18 - 6\sqrt{13}}{2\sqrt{13}}$$

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🔹 Step 4 — Compute $(\beta - \alpha)^2$

$$\beta - \alpha = \frac{(18 - 6\sqrt{13}) - (12 - 6\sqrt{13})}{2\sqrt{13}} = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}}$$

Now square it:

$$(\beta - \alpha)^2 = \frac{9}{13}$$

So,

$$m = 9,\quad n = 13$$

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✅ Final Answer

$$m+n = 9 + 13 = \boxed{22}$$

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