Question

The equation of the circle whose center is (0, 3) and radius is length 4 is:

x² + (y - 3)² = 8

x² + (y - 3)² = 16

x² + (y + 3)² = 16

Answer 1

The equation of a circle

`(x-xc)^2+(y-yc)^2=r^2`
defines a circle centred at (xc,yc) with radius r.

So with (xc,yc)=(0,3), and r=4
the equation becomes

`(x-0)^2+(y-3)^2=4^2`

`x^2+(y-3)^2=16`