Mathematics · Circles

JEE Main 2025 — 3 April, Evening Shift — Question 29

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

Answer: D

$2 r^{2}=26$

$r^{2}=13$

$\mathrm{Eq}^{\mathrm{n}}$ of circle is $x(x-4)+y(y-6)=0$

$k(k-4)+3 k(3 k-6)=0$

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

$10 k^{2}=22 k$

$10 k=22$

$\therefore \quad 10 k+r^{2}=35$

Answer key and solution verified before publishing.

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