JEE Main 2022 (Online) 30th June Morning ShiftQuestion 1
Consider three circles :
$C_1:x^{2}+y^{2}=r^{2}$
$C_2:(x-1)^{2}+(y-1)^{2}=r^{2}$
$C_3:(x-2)^{2}+(y-1)^{2}=r^{2}$
If a line $L:y=mx+c$ be a common tangent to $C_1$, $C_2$ and $C_3$ such that $C_1$ and $C_3$ lie on one side of line $L$ while $C_2$ lies on other side, then the value of $20\left(r^{2}+c\right)$ is equal to :
Circle
Solution
Signed distances of centres $(0,0),(1,1),(2,1)$ from $mx-y+c=0$ are proportional to $c,\ m-1+c,\ 2m-1+c$. $C_1,C_3$ same side with equal distance: $c=2m-1+c\Rightarrow m=\frac12$. $C_2$ opposite: $m-1+c=-c\Rightarrow c=\frac14$. $r^2=\frac{c^2}{1+m^2}=\frac{1/16}{5/4}=\frac1{20}$. $20(r^2+c)=1+5=6$. Option (D).