Question

What is the eqaition of this circle?

Answer 1

Given the center point:

(-3, 2)

Let's find the equation of the circle with the center above which is tangent to the y-axis.

Apply the equation of a circle:

`(x-h)^2+(y-k)^2=r^2`

Where (h, k) is the center of the circle and the radius is r.

Since it is tangent to the y-axis, the radius of the circle is the x-coordinate of the center point.

We have:

(h, k) ==> (-3, 2)

r = -3

To write the equation of the circle, we have:

`\begin{gathered} (x--3)^2+(y-2)^2=-3^2 \\ \\ (x+3)^2+(y-2)^2=9 \end{gathered}`

Thus, the equation of the circle is:

`(x+3)^2+(y-2)^2=9`

ANSWER:

`(x+3)^2+(y-2)^2=9`