Question

If the radius of the circle $x^2+y^2-18x-12y+k=0$ be $11$ then k=

• A. 34
• B. 4
• C. -4
• D. 49

Answer 1

The correct option is C -4

The equation of circle is $x^2+y^2-18x-12y+k=0$.

Comparing it with $x^2+y^2+2gx+2fy+c=0$ we get,

$2g=-18,2f=-12$ and $c=k$

$\therefore$ $g=-9,f=-6$ and $c=k$

Center $=(-g,-f)=(9,6)$

Radius of a circle $\left(r\right)=11$ [ Given ]

We know,

$\Rightarrow$ $r=\sqrt{g^2+f^2-c}$

$\Rightarrow$ $11=\sqrt{(-9{)}^2+(-6{)}^2-k}$

$\Rightarrow$ $11=\sqrt{81+36-k}$

$\Rightarrow$ $11=\sqrt{117-k}$

$\Rightarrow$ $121=117-k$

$\Rightarrow$ $4=-k$

$\therefore$ $k=-4$