JEE Main 2021 (Online) 3rd August Evening ShiftQuestion 1
The circles, $x^{2}+y^{2}+2x+4y-4=0$ and $x^{2}+y^{2}-4x+4y+k=0$ touch each other internally. If the point of contact of these circles is $(a,b)$, then $5\left(a^{2}+b^{2}\right)$ is equal to $\_\_\_\_\_\_$ .
Circle
Answer$100$
Solution
$C_1$: centre $(-1,-2)$, $r_1=3$. $C_2$: centre $(2,-2)$, $r_2=\sqrt{8-k}$. Distance of centres $=3=|r_2-r_1|$, so $r_2=6$ ($r_2=0$ rejected), $k=-28$. Contact point lies on line of centres, on $C_1$ farthest from $C_2$'s centre: $(-1,-2)+3(-1,0)=(-4,-2)$. $5(a^2+b^2)=5(16+4)=100$.