Question

Find the centre and radius of the circle $\sqrt{1+m^2}(x^2+y^2)-2cx-2mcy=0$

Answer 1

The equation of the circle is given as $\sqrt{1+m^2}\left(x^2+y^2\right)-2cx-2mcy=0$.

$\left(x^2+y^2\right)-\frac{2cx}{\sqrt{1+m^2}}-\frac{2mcy}{\sqrt{1+m^2}}=0$

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,

$g=-\frac{c}{\sqrt{1+m^2}}$ and $f=-\frac{mc}{\sqrt{1+m^2}}$

$C=\left(\frac{c}{\sqrt{1+m^2}},\frac{mc}{\sqrt{1+m^2}}\right)$

And the radius is,

$r=\sqrt{{\left(-\frac{c}{\sqrt{1+m^2}}\right)}^2+{\left(-\frac{mc}{\sqrt{1+m^2}}\right)}^2}$

$=\sqrt{\frac{c^2}{1+m^2}+\frac{m^2{c}^2}{1+m^2}}$

$=\sqrt{\frac{c^2\left(1+m^2\right)}{1+m^2}}$

$=c$

Therefore, the center of the circle is $\left(\frac{c}{\sqrt{1+m^2}},\frac{mc}{\sqrt{1+m^2}}\right)$and radius of the circle is $c$.