JEE Main 2025 Apr 7 Shift 1, Mathematics Q9: Involving Two or More Circles

JEE Main2025Apr 7, Shift 1Mathematics

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 $(β - α)^{2} = \frac{m}{n}$ , $g cd (m, n) = 1$ , then $m + n$ is equal to :

A
$9$
B
$13$
C
$22$
D
$31$

Solution

Locate the circles. $C_{1}$ has centre $A = (- 3, - 3)$ and radius $r_{1} = 3$ . Let $C_{2}$ have centre $B = (1, 3)$ and radius $r_{2}$ .

Distance between centres: $A B = 4^{2} + 6^{2} = 52 = 213$ .

For external tangency: $A B = r_{1} + r_{2} \Rightarrow r_{2} = 213 - 3$ .

Point of tangency $P = (α, β)$ divides $A B$ internally in the ratio $r_{1} : r_{2}$ . Using the section formula:

$α = \frac{r _{1} \cdot 1 + r _{2} \cdot ( - 3 )}{r _{1} + r _{2}} = \frac{3 - 3 ( 2 13 - 3 )}{2 13} = \frac{3 - 6 13 + 9}{2 13} = \frac{12 - 6 13}{2 13} = \frac{6 - 3 13}{13}$

$β = \frac{r _{1} \cdot 3 + r _{2} \cdot ( - 3 )}{r _{1} + r _{2}} = \frac{9 - 3 ( 2 13 - 3 )}{2 13} = \frac{18 - 6 13}{2 13} = \frac{9 - 3 13}{13}$

Compute $β - α$ :

$β - α = \frac{( 9 - 3 13 ) - ( 6 - 3 13 )}{13} = \frac{3}{13}$

$(β - α)^{2} = (\frac{3}{13})^{2} = \frac{9}{13}$

So $m = 9$ , $n = 13$ , and $g cd (9, 13) = 1$ .

$∴ m + n = 22 \Rightarrow$ Option (3).

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JEE Main 2025 Apr 7 Shift 1, Mathematics Q9: Involving Two or More Circles

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