Question

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

Answer 1

Let centre of circle be $(\alpha ,\beta )$

$\therefore (\alpha -1{)}^2+(\beta -0{)}^2=(\alpha -0{)}^2+(\beta -1{)}^2$

$\therefore {\alpha }^2-2\alpha +1+{\beta }^2={\alpha }^2+{\beta }^2-2\beta +1$

$\Rightarrow \alpha =\beta$

And

$(\alpha -1{)}^2+(\beta -0{)}^2=(\alpha -1{)}^2+(\beta +2{)}^2$

$\Rightarrow {\beta }^2={\beta }^2+4\beta +4$

$\Rightarrow \beta =-1=\alpha$

$\therefore$ Radius of circle = $\sqrt{(\alpha -1{)}^2+(\beta -0{)}^2}$

$=\sqrt{(-1-1{)}^2+(-1{)}^2}$

$=\sqrt{5}$

$\therefore$ equation of circle :- $(x+1{)}^2+(y+1{)}^2=5$