Question

Find the equation of 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 (r) of the circle is the distance between the point (2,2) and (4,5).

$r=\sqrt{(2-4{)}^2+(2-5{)}^2}=\sqrt{13}$

Thus, the equation of the circle is

$(x-h{)}^2+(y-k{)}^2=r^2$

$(x-2{)}^2+(y-2{)}^2=13$

$x^2-4x+4+y^2-4y+4=13$

$x^2+y^2-4x-4y-5=0$