Question

If $d_1$ and $d_2$ are the longest and the shortest distances of the point $P(-7,2)$ from the circle $x^2+y^2-10x-14y-51=0,$ then the value of $d_1^2+d_2^2$ is

• A. 588.0
• B. 588
• C. 588.00

Answer 1

Answer: (A), (B), (C)

Given circle $S\equiv x^2+y^2-10x-14y-51=0$, $P$ is $(-7,2)$
$S_1=(-7{)}^2+2^2-10(-7)-14(2)-51=44>0$
$\Rightarrow P$ lies outside the circle.
Centre, $C=(-g,-f)=(5,7)$
Radius, $r=\sqrt{g^2+f^2-c}=\sqrt{25+49+51}=\sqrt{125}$
$CP=\sqrt{{12}^2+5^2}=13$
$\therefore d_1=CP+r=13+\sqrt{125}$
$d_2=CP-r=13-\sqrt{125}$
$\therefore d_1^2+d_2^2=(13+\sqrt{125}{)}^2+(13-\sqrt{125}{)}^2=2(169+125)=588$