Question

If the power of point $(2,1)$ with respect to the circle $2x^2+2y^2-8x-6y+k=0$ is positive, then

• A. $12<k<\infty$
• B. $-12<k<12$
• C. $12<k<\frac{25}{2}$
• D. $12<k<25$

Answer 1

The correct option is C $12<k<\frac{25}{2}$

$2x^2+2y^2-8x-6y+k=0$
$\Rightarrow x^2+y^2-4x-3y+\frac{k}{2}=0$
Radius $=\sqrt{4+\frac{9}{4}-\frac{k}{2}}=\sqrt{\frac{25-2k}{4}}$
Clearly, $\frac{25-2k}{4}>0$
$\Rightarrow k<\frac{25}{2}\cdots \left(1\right)$

The power of the point $(2,1)$ with respect to the circle $2x^2+2y^2-8x-6y+k=0$ is
$S_1=2(2{)}^2+2(1{)}^2-8\left(2\right)-6\left(1\right)+k$
$\Rightarrow S_1=-12+k$
Given $S_1>0$
Then $-12+k>0$
$\Rightarrow k>12\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$12<k<\frac{25}{2}$