Question

Find the equation of the circle which passes through the point $(1,1)$ & center at the point $(2,3)$ on it.

Answer 1

Center of circle is $(2,3)$

The point passing through is $(1,1)$

Let the radius be $r$

The equation of circle is given as $(x-2{)}^2+(y-3{)}^2=r^2$

It passes through $(1,1)⟹(1-2{)}^2+(1-3{)}^2=r^{2}1+4=r^2{r}^2=5$

So the equation of circle is $(x-2{)}^2+(y-3{)}^2=5x^2-4x+4+y^2-6y+9=5x^2+y^2-4x-6y+8=0$