Question

A circle has its center in Quadrant I. It is tangent to the y-axis at (0,2) and its radius is 3. What is the equation of the circle? Explain the process you use to find the equation.

Answer 1

we know that

the equation of a circle is
(x-h)²+(y-k)²=r²
r=3 units
where the center is the point (h,k)

if the circle is tangent to the y-axis at (0,2) and its center in Quadrant I
then
the center is the point (0+3,2)--------> (3,2)
therefore

the answer is
(x-3)²+(y-2)²=9

see the attached figure