Question

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

Answer 1

The equation of circle is
$(x-h{)}^2+(y-k{)}^2=r^2$
Since teh circle passes through point (4,5) and co-orcinates of centre are (2,2)
$\therefore \text{radius of circle}=\sqrt{(4-2{)}^2+(5-2{)}^2}=\sqrt{4+9}=\sqrt{13}\text{Now the equation of required circle is}(x-2{)}^2+(y-2{)}^2=(\sqrt{13}{)}^2\Rightarrow x^2+4-4x+y^2+4-4y=13\Rightarrow x^2+y^2-4x-4y-5=0.$