Center (−9,−4/3)
and tangent to the y-axis

To find the equation of a circle that is centered at \((-9, -\frac{4}{3})\) and is tangent to the y-axis, follow these steps:

### Step 1: Identify the center and radius
- The center of the circle is given as \((-9, -\frac{4}{3})\).
- Since the circle is tangent to the y-axis, the radius of the circle is the horizontal distance from the center to the y-axis.

### Step 2: Calculate the radius
- The distance from the center to the y-axis is the absolute value of the x-coordinate of the center, which is \(|-9| = 9\).

### Step 3: Write the equation of the circle
- The general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.
- Plugging in the values, we get:

\[
(x + 9)^2 + \left(y + \frac{4}{3}\right)^2 = 9^2
\]

\[
(x + 9)^2 + \left(y + \frac{4}{3}\right)^2 = 81
\]

This is the equation of the circle with the given center and tangent to the y-axis.

Would you like further details on this problem? Here are some related questions:

1. How do you calculate the distance between two points?
2. What is the significance of a circle being tangent to a line?
3. How can you find the points of intersection between a circle and a line?
4. How does the general equation of a circle change if it is tangent to the x-axis instead?
5. Can you derive the equation of a circle given a different center and radius?

**Tip:** The distance formula is essential when working with geometric shapes on the coordinate plane, particularly when determining radii and distances between points.

Discover how to find the equation of a circle centered at (-9, -4/3) and tangent to the y-axis. Step-by-step solution with the application of the distance formula. Suitable for Grades 9-12.

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