Let be the circle in the third quadrant of radius 3, that touches both coordinate axes. Let be the circle with centre (1, 3) that touches externally at the point . If , gcd (m, n) = 1, then m + n is equal to [2025]

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Solution

Q.
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 $(α, β)$ . If ${(β - α)}^{2} = \frac{m}{n}$ , gcd (m, n) = 1, then m + n is equal to [2025]

1 31

2 13

3 9

4 22

Ans.
(4)

The equation of circle $C_{1}, {(x + 3)}^{2} + {(y + 3)}^{2} = 3^{2}$

The centres has $C_{1}$ and $C_{2}$ are A(–3, –3) and B(1, 3)

$A B = \sqrt{16 + 36} = 2 \sqrt{13}$

So, the radius are $r_{1} = 3$ and $r_{2} = 2 \sqrt{13} - 3$

The point $P (α, β)$

$α = \frac{r_{1} (1) + r_{2} (- 3)}{r_{1} + r_{2}}, β = \frac{r_{1} (3) + r_{2} (- 3)}{r_{1} + r_{2}}$

$α = \frac{3 - 3 (2 \sqrt{13} - 3)}{2 \sqrt{13}}, β = \frac{3 (3) + (2 \sqrt{3} - 3) (- 3)}{2 \sqrt{13}}$

$\Rightarrow α = \frac{12 - 6 \sqrt{3}}{2 \sqrt{3}}, β = \frac{18 - 6 \sqrt{3}}{2 \sqrt{3}}$

Now, ${(β - α)}^{2} = {(\frac{6}{2 \sqrt{13}})}^{2} = \frac{36}{52} = \frac{9}{13}$ $(Given, {(β - α)}^{2} = \frac{m}{n})$

So, m = 9, n = 13.

So, m + n = 22.

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