Question

Find the Centre and radius

Answer 1

The equation of the given circle is

$\quad\:\mathsf{3x^{2}+3y^{2}-18x+6y+7=0}$

$:\to \mathsf{x^{2}+y^{2}-6x+2y+\frac{7}{3}=0}$

$:\to \small{\mathsf{(x^{2}-6x+9)+(y^{2}+2y+1)=9+1-\frac{7}{3}}}$

$:\to \mathsf{(x-3)^{2}+(y+1)^{2}=\left(\sqrt{\frac{23}{3}}\right)^{2}}$

Comparing it with the general equation of circle

$\quad\mathsf{(x-a)^{2}+(y-b)^{2}=r^{2}}$

where (a, b) is the centre of the circle and r is it's radius, we get the centre to be at (3, - 1) and radius to be $\mathsf{\left(\sqrt{\frac{23}{3}}\right)}$ units