JEE Main 2025 Apr 3 Shift 2, Mathematics Q3: Involving Two or More Circles

JEE Main2025Apr 3, Shift 2Mathematics

If the four distinct points $(4, 6)$ , $(- 1, 5)$ , $(0, 0)$ and $(k, 3 k)$ lie on a circle of radius $r$ , then $10 k + r^{2}$ is equal to

Solution

$m_{1} m_{2} = - 1$

$∴$ The right-angle circle is

$(x - 4) (x - 0) + (y - 6) (y - 0) = 0$

$\Rightarrow x^{2} + y^{2} - 4 x - 6 y = 0$

Since $(k, 3 k)$ lies on the circle,

$k^{2} + 9 k^{2} - 4 k - 18 k = 0$

$\Rightarrow 10 k^{2} - 22 k = 0$

$\Rightarrow k = 0, \frac{11}{5}$

$k = 0$ is not possible.

$∴ k = \frac{11}{5}$

Also,

$r = 2^{2} + 3^{2} = 13$

$∴ 10 k + r^{2} = 10 (\frac{11}{5}) + (13)^{2} = 22 + 13 = 35$

$35$

Concept behind this question

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