Question

Find the diameters of centre of circle passing through the point $(0,0)(-2,1)$ and $(-3,2)$.

Answer 1

$\begin{array}{c}S\tan dardequationcircleis\\ x^2+y^2+2gx+2fy+c=0\\ \Rightarrow points\left(0,1\right)\left(-2,1\right)and\left(-3,2\right)satisfiestocirclec=0\\ \Rightarrow 4+1-4g+2f+0=0\\ \Rightarrow 2f-4g=-5\to \left(i\right)\\ andput\left(-3,2\right)\\ \Rightarrow 9+4-6g+4f=0\\ \Rightarrow 4f-6g+4f=0\\ \Rightarrow 4f-6g=-13\\ 2f-3g=-13\to \left(ii\right)\\ bysolvingequation\left(i\right)and\left(ii\right)weget\\ \Rightarrow \frac{f}{-26+15}=\frac{g}{\left(-13+10\right)}=\frac{1}{-6+8}\\ \frac{f}{-11}=\frac{g}{-3}=\frac{1}{2}\\ \Rightarrow f=\frac{-11}{2},g=\frac{-3}{2}\\ hence,radiusofcircle=\sqrt{f^2+g^2-c}\\ radius=\sqrt{{\left(\frac{-11}{2}\right)}^2+{\left(\frac{-3}{2}\right)}^2}=\sqrt{\frac{13}{4}}=\sqrt{\frac{65}{2}}\end{array}$