Question

The nearest point on the circle $x^2+y^2-6x+4y-12=0$ from the point $P(-5,4)$ is $Q(\alpha ,\beta ),$ then the value of $\alpha +\beta$ is

• A. $0$
• B. $4$
• C. $2$
• D. $7$

Answer 1

The correct option is A $0$

Given circle: $S\equiv x^2+y^2-6x+4y-12=0$
Centre, $C\equiv (3,-2)$
radius$,r=\sqrt{9+4+12}=5$
Point. $P\equiv (-5,4)$
$S_1=25+16+30+16-12=75>0$
$\Rightarrow P$ lies outside the circle
$CP=\sqrt{64+36}=10$
Let $Q$ be the required point
$\therefore PQ=CP-r=5$
$\Rightarrow Q$ is the midpoint of $CP$
$\Rightarrow Q=(\frac{-5+3}{2},\frac{4-2}{2})=(-1,1)$
$\therefore \alpha =-1,\beta =1\Rightarrow \alpha +\beta =0$