Question

Let $x,y$ be real variables satisfying the $x^2+y^2+8x-10y-40=0$. Let $a=max\left\{\sqrt{(x+2{)}^2+(y-3{)}^2}\right\}$ and $b=min\left\{\sqrt{(x+2{)}^2+(y-3{)}^2}\right\}$, then

• A. $a+b=18$
• B. $a+b=\sqrt{2}$
• C. $a-b=4\sqrt{2}$
• D. $a.b=73$

Answer 1

Answer: (A), (C), (D)

$a.b=73$

$x^2+y^2+8x-10y-40=0$
The centre of the circle is $(-4,5)$.
Its radius is $9$.

Distance of the center $(-4,5)$ from the point $(-2,3)$ is $\sqrt{4+4}=2\sqrt{2}$. Therefore,
$a=2\sqrt{2}+9$ and $b=-2\sqrt{2}+9$
$\therefore a+b=18$
$a-b=4\sqrt{2}$
$a.b=81-8=73$