Question

Find the equation of the circle whose centre lies on the positive direction of y-axis at a distance 6 from the origin and whose radius is 4.

Answer 1

The general equation of circle, with centre (a, b) and radius r in
$(x-a{)}^2+(y-b{)}^2=r^2\dots \left(A\right)$
Now,
According to the question centre = (0, 6) and radius = 4
$\therefore$ From (A)
$\Rightarrow (x-0{)}^2+(y-6{)}^2=4^2\Rightarrow x^2+y^2—12y+20=0$