Question

The point $P(10,7)$ lies outside the circle $C:x^2+y^2-4x-2y-20=0$. $l,m$ denote respectively the least and greatest distance of P from the points on the circle $C$. Then first digit of value of ${(9l+2m)}^2$ is equal to

Answer 1

Centre of the given circle is $A\equiv (2,1)$ and radius, $r=\sqrt{2^2+1^2+20}=5$
$\Rightarrow CA=\sqrt{(10-2{)}^2+(7-1{)}^2)}=10$
$\therefore l=CA-r=5$ and $m=CA+r=15$
Hence, $(9l+2m{)}^2=5625$