Question

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

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

Answer 1

The correct option is A $a+b=18$

$x^2+y^2+8x-10y+40=0$
$\Rightarrow (x+4{)}^2+(y-5{)}^2=1$
Let $x=\cos \theta -4,y=\sin \theta +5$
Now, $(x+2{)}^2+(y-3{)}^2$
$=(\cos \theta -2{)}^2+(\sin \theta +2{)}^2$
$=9+4(\sin \theta -\cos \theta )$
$a=max\{9+4(\sin \theta -\cos \theta \left)\right\}=9+4\sqrt{2}$
$b=min\{9+4(\sin \theta -\cos \theta \left)\right\}=9-4\sqrt{2}$
$a+b=18$
$a-b=8\sqrt{2}$
$ab=49$