Write the standard form of the equation of the circle described below.

Center 

​(0,negative 2−2​) 

passes through the point 

​(negative 4−4​,negative 5−5​)

The standard form of the equation of a circle is:

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

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

Given:
- Center: \((0, -2)\)
- A point on the circle: \((-4, -5)\)

### Step 1: Find the radius
The radius \(r\) is the distance from the center of the circle to any point on the circle. We can use the distance formula:

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

Substituting the coordinates of the center \((0, -2)\) and the point \((-4, -5)\):

\[
r = \sqrt{(0 - (-4))^2 + (-2 - (-5))^2}
\]
\[
r = \sqrt{(4)^2 + (3)^2}
\]
\[
r = \sqrt{16 + 9} = \sqrt{25} = 5
\]

### Step 2: Write the equation of the circle
Now that we know the radius is \(5\), we can plug the values into the standard form:

\[
(x - 0)^2 + (y - (-2))^2 = 5^2
\]
\[
x^2 + (y + 2)^2 = 25
\]

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

\[
x^2 + (y + 2)^2 = 25
\]

Would you like further details or explanations?

Here are 5 related questions to expand on this:
1. How do you find the equation of a circle if given two points on the circle?
2. Can the equation of a circle be written in general form?
3. How do you determine if a point lies inside, on, or outside a circle?
4. What happens to the equation if the circle is centered at the origin?
5. How do you find the tangent to a circle at a given point?

**Tip:** The radius is always a positive value, representing the distance from the center to any point on the circle.

Write the standard form of the equation of the circle described below. Center (0, -2) passes through the point (-4, -5).

Learn how to find the standard form of the equation of a circle with center (0, -2) and a point (-4, -5) on the circle. This solution involves using the distance formula to calculate the radius and writing the equation of the circle. Suitable for students in Grades 9-10.

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