Find the standard form of the equation for the circle with the following properties.

Center (2,−11)(2,−11), passes through (−2,−5)


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

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

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

Given:
- Center \((h, k) = (2, -11)\)
- The circle passes through the point \((-2, -5)\)

### Step 1: Find the radius \(r\)

We can calculate the radius by finding the distance between the center \((2, -11)\) and the point on the circle \((-2, -5)\) using the distance formula:

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

Substituting the points \((x_1, y_1) = (2, -11)\) and \((x_2, y_2) = (-2, -5)\):

\[
r = \sqrt{((-2) - 2)^2 + ((-5) - (-11))^2}
\]
\[
r = \sqrt{(-4)^2 + (6)^2}
\]
\[
r = \sqrt{16 + 36} = \sqrt{52}
\]
\[
r = 2\sqrt{13}
\]

### Step 2: Write the standard form of the equation

Now that we know the radius \(r = 2\sqrt{13}\), we can substitute the values into the equation:

\[
(x - 2)^2 + (y + 11)^2 = (2\sqrt{13})^2
\]
\[
(x - 2)^2 + (y + 11)^2 = 52
\]

Thus, the standard form of the equation of the circle is:

\[
(x - 2)^2 + (y + 11)^2 = 52
\]

Would you like more details or have any other questions?

Here are 5 related questions to expand on this:
1. How do you convert the equation of a circle from standard form to general form?
2. How would the equation change if the radius were doubled?
3. What are some other methods for finding the radius of a circle?
4. Can you determine the diameter of the circle from this equation?
5. How do you graph a circle given its standard form?

**Tip:** When dealing with equations of circles, always ensure the radius is squared on the right side of the equation.

Find the standard form of the equation for the circle with the following properties: Center (2,−11), passes through (−2,−5).

Discover how to find the standard form of a circle's equation with center (2,−11) and passing through the point (−2,−5). This problem uses the distance formula and the equation of a circle to solve for the radius and equation. Suitable for students in Grades 9-11.

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