Question

Find the equation of a circle with centre $(2,2)$ and passes through the point $(4,5)$.

Answer 1

The centre of the circle is given as $(h,k)=(2,2)$
Since the circle passes through point $(4,5)$ the radius $\left(r\right)$ of the circle is the distance between the point $(2,2)$ and $(4,5)$
$\therefore r=\sqrt{{(2-4)}^2+{(2-5)}^2}=\sqrt{{(-2)}^2+{(-3)}^2=}$$\sqrt{4+9}=\sqrt{13}$
Thus the equation of the circle is
${(x-h)}^2+{(y-k)}^2=r^2$
${(x-2)}^2+{(y-2)}^2={\left(\sqrt{13}\right)}^2$
$x^2-4x+4+y^2-4y+4=13$
$x^2+y^2-4x-4y-5=0$