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Let's work through the questions step by step based on the image you provided. I'll solve and explain each one.

### Exercises 118–119: Distance Between Each Pair of Points
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

**118. Points \((-2, 3)\) and \((3, -9)\):**
\[
d = \sqrt{(3 - (-2))^2 + (-9 - 3)^2} = \sqrt{(3 + 2)^2 + (-9 - 3)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
\]
The distance is **13**.

**119. Points \((-4, 3)\) and \((-2, 5)\):**
\[
d = \sqrt{(-2 - (-4))^2 + (5 - 3)^2} = \sqrt{(-2 + 4)^2 + (5 - 3)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83
\]
The distance is approximately **2.83**.

---

### Exercises 120–121: Midpoint of Each Line Segment
The midpoint \(M\) of a line segment with endpoints \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[
M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
\]

**120. Points \((2, 6)\) and \((-12, 4)\):**
\[
M = \left(\frac{2 + (-12)}{2}, \frac{6 + 4}{2}\right) = \left(\frac{-10}{2}, \frac{10}{2}\right) = (-5, 5)
\]
The midpoint is \((-5, 5)\).

**121. Points \((4, -6)\) and \((-15, 2)\):**
\[
M = \left(\frac{4 + (-15)}{2}, \frac{-6 + 2}{2}\right) = \left(\frac{-11}{2}, \frac{-4}{2}\right) = \left(-5.5, -2\right)
\]
The midpoint is \((-5.5, -2)\).

---

### Exercises 122–123: Equation of a Circle
The equation of a circle with center \((h, k)\) and radius \(r\) is:
\[
(x - h)^2 + (y - k)^2 = r^2
\]

**122. Center \((0, 0)\), \(r = 3\):**
\[
(x - 0)^2 + (y - 0)^2 = 3^2 \quad \Rightarrow \quad x^2 + y^2 = 9
\]
The equation is \(x^2 + y^2 = 9\).

**123. Center \((-2, 4)\), \(r = 6\):**
\[
(x - (-2))^2 + (y - 4)^2 = 6^2 \quad \Rightarrow \quad (x + 2)^2 + (y - 4)^2 = 36
\]
The equation is \((x + 2)^2 + (y - 4)^2 = 36\).

---

### Exercises 124–126: Center and Radius of Each Circle
To find the center and radius, we rewrite the given equations in standard form \((x - h)^2 + (y - k)^2 = r^2\).

**124. \(x^2 + y^2 = 1\):**
- Center: \((0, 0)\)
- Radius: \(r = \sqrt{1} = 1\)

**125. \((x + 2)^2 + (y - 3)^2 = 9\):**
- Center: \((-2, 3)\)
- Radius: \(r = \sqrt{9} = 3\)

**126. \(x^2 + y^2 - 4x + 2y - 4 = 0\):**
First, we complete the square:
\[
x^2 - 4x + y^2 + 2y = 4
\]
Completing the square for \(x\) and \(y\):
\[
(x^2 - 4x + 4) + (y^2 + 2y + 1) = 4 + 4 + 1 \quad \Rightarrow \quad (x - 2)^2 + (y + 1)^2 = 9
\]
- Center: \((2, -1)\)
- Radius: \(r = \sqrt{9} = 3\)

---

Let me know if you'd like further clarification, or if you have any questions. Here are some related questions:

1. How would you derive the distance formula from the Pythagorean theorem?
2. Can you generalize the midpoint formula for three-dimensional coordinates?
3. How do you complete the square to rewrite a general quadratic equation into circle form?
4. What is the relationship between the equation of a circle and conic sections?
5. How do you determine the domain and range of a circle's equation?

**Tip:** When solving problems involving circles, always rewrite the equation in standard form to easily identify the center and radius.

In Exercises 118-126, the user is asked to find the distance between points, the midpoint of line segments, the standard form of the equation of a circle, and the center and radius of circles.

This page provides solutions to exercises involving the distance between points, the midpoint formula, and the equation of a circle. Learn how to calculate the distance between two points, find the midpoint of a line segment, and write the equation of a circle in standard form. Suitable for students in Grades 9-11.

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