JEE 2025 — Mathematics PYQ

Let centre of circle = (a, 0)
radius = r
Distance of centre from line $x-y-3=0$ is $r$
∴ $r=\left|\frac{a-3}{\sqrt{2}}\right|$

Equation of circle,
$(x-a{)}^2+y^2={\left(\frac{a-3}{\sqrt{2}}\right)}^2$

∴ It passes through (-9, 4)
$2(a^2+18a+81)+16=(a^2-6a+9)$
$a^2+42a+185=0$
$(a+37)(a+5)=0$
∴ $a=-37,-5$

As,
$r=\left|\frac{a-3}{\sqrt{2}}\right|$
$r_1=\left|\frac{-37-3}{\sqrt{2}}\right|=20\sqrt{2}$
$r_2=\left|\frac{-5-3}{\sqrt{2}}\right|=4\sqrt{2}$
$|r_1^2-r_2^2|=|800-32|=768$