Solve for the Area, center, radius and standard equation using the endpoints of a diameter. Center (using midpoint formula) $$ \begin{array}{l} h=\frac{x_{1}+x_{2}}{2} \ k=\frac{y_{1}+y_{2}}{2} \end{array} $$ Area ofthe Clrele $$ = $$ un 13 Radius (use center as $$ \left(x_{1}, y_{1}\right) $$ And any point from the diameter will serve as $$ \left(x_{1}, y_{1}\right) $$ $$ r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} $$

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Sure! Let's go through the process of finding the **center**, **radius**, **area**, and the **standard equation** of a circle using the endpoints of its diameter. We'll use the provided formulas and illustrate each step with a hypothetical example.

### Given:
Suppose the endpoints of the diameter of a circle are:
- Point $$ A(x_1, y_1) $$
- Point $$ B(x_2, y_2) $$

### 1. **Finding the Center (Midpoint of the Diameter)**
The center $$(h, k)$$ of the circle is the midpoint of the diameter connecting points $$ A $$ and $$ B $$. Use the **midpoint formula**:

$$
\begin{cases}
h = \frac{x_1 + x_2}{2} \
k = \frac{y_1 + y_2}{2}
\end{cases}
$$

**Example:**
Let's say the endpoints are $$ A(2, 3) $$ and $$ B(8, 7) $$.

$$
\begin{align*}
h &= \frac{2 + 8}{2} = \frac{10}{2} = 5 \
k &= \frac{3 + 7}{2} = \frac{10}{2} = 5
\end{align*}
$$

So, the center is $$ (5, 5) $$.

### 2. **Calculating the Radius**
The **radius (r)** is half the length of the diameter. First, find the distance between points $$ A $$ and $$ B $$ using the **distance formula**:

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

Then, divide by 2 to get the radius:

$$
r = \frac{d}{2}
$$

**Continuing the Example:**

$$
\begin{align*}
d &= \sqrt{(8 - 2)^2 + (7 - 3)^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} \
r &= \frac{\sqrt{52}}{2} = \frac{2\sqrt{13}}{2} = \sqrt{13} \approx 3.606
\end{align*}
$$

### 3. **Writing the Standard Equation of the Circle**
The **standard equation** of a circle with center $$ (h, k) $$ and radius $$ r $$ is:

$$
(x - h)^2 + (y - k)^2 = r^2
$$

**Using Our Example:**

$$
(x - 5)^2 + (y - 5)^2 = (\sqrt{13})^2 \
(x - 5)^2 + (y - 5)^2 = 13
$$

### 4. **Calculating the Area of the Circle**
The **area (A)** of a circle is given by:

$$
A = \pi r^2
$$

**Using Our Example:**

$$
A = \pi (\sqrt{13})^2 = \pi \times 13 = 13\pi \approx 40.84 \text{ square units}
$$

### **Summary of Steps:**
1. **Center**: Use the midpoint formula with the endpoints of the diameter.
2. **Radius**: Calculate the distance between the endpoints and divide by 2.
3. **Standard Equation**: Plug the center coordinates and radius into $$(x - h)^2 + (y - k)^2 = r^2$$.
4. **Area**: Use $$ A = \pi r^2 $$.

### **Final Results from the Example:**
- **Center**: $$ (5, 5) $$
- **Radius**: $$ \sqrt{13} $$ units
- **Standard Equation**: $$ (x - 5)^2 + (y - 5)^2 = 13 $$
- **Area**: $$ 13\pi $$ square units

---

**Note:** Replace the example coordinates with your specific endpoints to perform these calculations for your particular circle.
To find the center, radius, area, and standard equation of a circle using the endpoints of its diameter: 1. **Center (Midpoint):** $$ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2} $$ 2. **Radius:** $$ r = \frac{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}{2} $$ 3. **Standard Equation:** $$ (x - h)^2 + (y - k)^2 = r^2 $$ 4. **Area:** $$ A = \pi r^2 $$ **Example:** For points $$ A(2, 3) $$ and $$ B(8, 7) $$: - **Center**: $$ (5, 5) $$ - **Radius**: $$ \sqrt{13} $$ units - **Standard Equation**: $$ (x - 5)^2 + (y - 5)^2 = 13 $$ - **Area**: $$ 13\pi $$ square units

Solve for the Area, center, radius and standard equation using the endpoints of a diameter.
Center (using midpoint formula)

Area ofthe Clrele un 13

Radius (use center as And any point from the diameter will serve as

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Solve for the Area, center, radius and standard equation using the endpoints of a diameter. Center (using midpoint formula) Area ofthe Clrele un 13 Radius (use center as And any point from the diameter will serve as