Find an equation of the circle with center at (3,-2) that is tangent to the y-axis in the form of (x-A)^2 + (y-B)^2 = C where A,B,C are constant.

We are asked to find the equation of a circle with a center at \( (3, -2) \) that is tangent to the y-axis. The general form of a circle's equation is:

\[
(x - A)^2 + (y - B)^2 = C
\]

where \( (A, B) \) is the center of the circle, and \( C \) is the square of the radius.

### Step 1: Identify the center and radius.
- The center of the circle is given as \( (3, -2) \), so \( A = 3 \) and \( B = -2 \).
- Since the circle is tangent to the y-axis, the distance from the center to the y-axis is the radius. The distance from \( (3, -2) \) to the y-axis is 3 units (because the x-coordinate of the center is 3).

Thus, the radius \( r = 3 \).

### Step 2: Write the equation of the circle.
The radius is squared in the equation of a circle, so:

\[
r^2 = 3^2 = 9
\]

Substituting \( A = 3 \), \( B = -2 \), and \( r^2 = 9 \) into the equation of the circle:

\[
(x - 3)^2 + (y + 2)^2 = 9
\]

This is the equation of the circle.

### Final Answer:
\[
(x - 3)^2 + (y + 2)^2 = 9
\]

Would you like further details or have any questions? 

Here are some related questions for deeper understanding:
1. How is the distance formula used to find the radius in this problem?
2. What is the general equation of a circle and what do each of the terms represent?
3. How would the equation change if the circle was tangent to the x-axis instead?
4. What happens to the equation if the center is moved to a different location?
5. How would the equation change if the circle's radius doubled?

**Tip:** The distance from a point to the y-axis is simply the absolute value of the x-coordinate of the point!

Learn how to find the equation of a circle with a center at (3, -2) that is tangent to the y-axis. This problem covers key geometric concepts such as the general equation of a circle, finding the radius, and using the distance formula. Suitable for students in Grades 9-11.

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