In the graph on the​ right, a line segment through the center of the circle intersects the circle at the points 

​(5​,5​) 

and 

​(7​, 

15​) 

as shown.

**a.** Find the coordinates of the​ circle's center.

**b.** Find the radius of the circle.

**c.** Use your answers from parts​ (a) and​ (b) to write the standard form of the​ circle's equation.


To solve this problem, we need to address the three parts sequentially:

### Part a: Find the coordinates of the circle's center
Given that the line segment passes through the center of the circle and intersects the circle at points \( (5, 5) \) and \( (7, 15) \), the center of the circle is located at the **midpoint** of these two points.

The midpoint formula is:

\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

Here, \( (x_1, y_1) = (5, 5) \) and \( (x_2, y_2) = (7, 15) \). Let's plug in the values:

\[
M = \left( \frac{5 + 7}{2}, \frac{5 + 15}{2} \right) = \left( \frac{12}{2}, \frac{20}{2} \right) = (6, 10)
\]

Thus, the coordinates of the center of the circle are \( (6, 10) \).

### Part b: Find the radius of the circle
The radius of the circle is the distance from the center of the circle to any point on the circumference. We can use the distance formula to find the distance between the center \( (6, 10) \) and one of the given points on the circle, say \( (5, 5) \).

The distance formula is:

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

Here, \( (x_1, y_1) = (6, 10) \) and \( (x_2, y_2) = (5, 5) \). Plugging in the values:

\[
d = \sqrt{(5 - 6)^2 + (5 - 10)^2} = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26}
\]

Thus, the radius \( r \) of the circle is \( \sqrt{26} \).

### Part c: Write the standard form of the circle's equation
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.

From part (a), we know the center is \( (6, 10) \), and from part (b), we know the radius is \( \sqrt{26} \), so \( r^2 = 26 \). Plugging these values into the standard form:

\[
(x - 6)^2 + (y - 10)^2 = 26
\]

This is the standard form of the equation of the circle.

---

Would you like more details on any part of this? Here are some related questions to consider:

1. How is the midpoint of a line segment calculated in general?
2. Can the distance formula be derived from the Pythagorean theorem?
3. What would the equation look like if the center of the circle was not on the origin?
4. How can you find the equation of a circle if given the diameter endpoints?
5. How does changing the radius affect the graph of a circle?

**Tip:** When writing the equation of a circle, always ensure you carefully identify the center and the radius, as they directly impact the equation's form.

In the graph on the right, a line segment through the center of the circle intersects the circle at the points (5,5) and (7, 15) as shown.

**a.** Find the coordinates of the circle's center.

**b.** Find the radius of the circle.

**c.** Use your answers from parts (a) and (b) to write the standard form of the circle's equation.

Learn how to find the center, radius, and standard equation of a circle given two points on the circumference. This problem covers key geometric concepts like the midpoint formula, distance formula, and the equation of a circle. Suitable for Grades 9-10.

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