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To find the correct equation of the circle, we use the standard form of the equation of a circle:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

Where:
- \((h, k)\) is the center of the circle.
- \(r\) is the radius.

From the problem:
- The center of the circle is \((5, 2)\), so \(h = 5\) and \(k = 2\).
- A point on the circle is \((5, 9)\), which we will use to find the radius.

### Step 1: Calculate the radius
The radius \(r\) is the distance between the center \((5, 2)\) and the point on the circle \((5, 9)\). The formula for the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[
r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Substitute the values:

\[
r = \sqrt{(5 - 5)^2 + (9 - 2)^2} = \sqrt{0 + 49} = \sqrt{49} = 7
\]

Thus, the radius is 7, and \(r^2 = 49\).

### Step 2: Write the equation of the circle
Now that we know the center and radius, we can substitute into the general equation of a circle:

\[
(x - 5)^2 + (y - 2)^2 = 49
\]

### Step 3: Find the correct option
From the answer choices:
- The correct answer is **C**: \((x - 5)^2 + (y - 2)^2 = 49\).

Would you like further details or clarification? Here are some related questions you can explore:

1. How do you find the equation of a circle given its diameter?
2. What is the relationship between the radius and the circumference of a circle?
3. How can the equation of a circle be derived from its general form?
4. How does changing the radius affect the graph of the circle?
5. What happens if the center of the circle is shifted to a different point?

**Tip**: When working with equations of circles, always verify the center and radius by substituting points from the circle to avoid errors!

A circle in the coordinate plane has its center at (5, 2). A point on the circle is found at (5, 9). Which of the following is an equation of the circle?

Find the equation of a circle with center at (5, 2) and a point on the circle at (5, 9). This problem uses the standard form of the equation of a circle and the distance formula to calculate the radius. Suitable for Grades 9-11.

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