Question

If x and y are real variables satisfying

$x^2+y^2+8x-10y+40=0$ and

$2a=max.\left[\right(x+2{)}^2+(y-3{)}^2]$;

$2b=min.\left[\right(x+2{)}^2+(y-3{)}^2]$ then $a+b$ is

Answer 1

Centre of given circle be $C(–4,5)$ and its radius $r=1$. Let the distance between two point A and B be $d\left(AB\right)$ so $2a=max{\left\{d\right(PQ\left)\right\}}^2$ where $P=(x,y)$ and $Q=(-2,3)$

= $\left[d\right(CQ)+r{]}^2={(\sqrt{(4-2{)}^2+(5-3{)}^2}+1)}^2$

= ${(2\sqrt{2}+1)}^2$

and $2b=min{\left\{d\right(PQ\left)\right\}}^2=\left[d\right(CQ)-r{]}^2$

= $[\sqrt{(4-2{)}^2+(5-3{)}^2}-1{]}^2$

= ${(2\sqrt{2}-1)}^2$

so $2(a+b)=(2\sqrt{2}+1{)}^2+(2\sqrt{2}-1{)}^2$

=$2(8+1)=18\Rightarrow a+b=9$