Question

Find the centre and radius of the circle $x^2+y^2+6x+8y-96=0$.

Answer 1

The equation of the circle is given as $x^2+y^2+6x+8y-96=0$.

Comparing the given equation with general equation of circle,

$x^2+y^2+2gx+2fy+c=0$

The center of circle is $C=\left(-g,-f\right)$ and the radius of circle is $r=\sqrt{g^2+f^2-c}$.

Therefore, from the given equation,

$2g=6,2f=8$

$g=3,f=4$

$C=\left(-3,-4\right)$

And the radius is,

$r=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2-\left(-96\right)}$

$=\sqrt{9+16+96}$

$=\sqrt{121}$

$=11$

Therefore, the center of the circle is $\left(-3,-4\right)$ and radius of the circle is $11$.