Question

Find the equation of the circle with centre at $(2,-1)$ and which passes through the point $(3,6)$.

Answer 1

As we know that equation of circle is given as-

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

Whereas, $(h,k)$ is the centre of circle and $r$ is the radius of circle.

Centre of circle $=(2,-1)\left(\text{Given}\right)$

Since the circle passes through the point $(3,6)$.

$\therefore$ Radius of circle $=\sqrt{{(2-3)}^2+{(-1-6)}^2}=\sqrt{1+49}=\sqrt{50}$

Therefre,

Equation of circle $\Rightarrow {(x-2)}^2+{(y-(-1))}^2={\left(\sqrt{50}\right)}^2\Rightarrow {(x-2)}^2+{(y+1)}^2=50$