Writing Circle Equations worksheet with ten problems to find the standard form equation of a circle using various given conditions.

Problem: Writing Circle Equations

We are tasked with writing the standard form equation of a circle for various given conditions. The standard form of a circle's equation is:

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

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

Let's solve each problem step by step.

---

1) Center: \((9, -8)\), Tangent to \(x = -1\)

#### Step 1: Identify the center
The center of the circle is given as \((h, k) = (9, -8)\).

#### Step 2: Determine the radius
The circle is tangent to the vertical line \(x = -1\). The distance from the center \((9, -8)\) to the line \(x = -1\) is the radius \(r\). This distance is calculated as:
\[
r = |9 - (-1)| = |9 + 1| = 10
\]

#### Step 3: Write the equation
Using the center \((9, -8)\) and radius \(r = 10\), the equation of the circle is:
\[
(x - 9)^2 + (y + 8)^2 = 10^2
\]
\[
(x - 9)^2 + (y + 8)^2 = 100
\]

#### Final Answer:
\[
\boxed{(x - 9)^2 + (y + 8)^2 = 100}
\]

---

2) Center: \((1, 9)\), Circumference: \(12\pi\)

#### Step 1: Identify the center
The center of the circle is given as \((h, k) = (1, 9)\).

#### Step 2: Determine the radius
The circumference of a circle is given by:
\[
C = 2\pi r
\]
We are given \(C = 12\pi\). Solving for \(r\):
\[
12\pi = 2\pi r \implies r = \frac{12\pi}{2\pi} = 6
\]

#### Step 3: Write the equation
Using the center \((1, 9)\) and radius \(r = 6\), the equation of the circle is:
\[
(x - 1)^2 + (y - 9)^2 = 6^2
\]
\[
(x - 1)^2 + (y - 9)^2 = 36
\]

#### Final Answer:
\[
\boxed{(x - 1)^2 + (y - 9)^2 = 36}
\]

---

3) Center: \((2, -3)\), Point on the Circle: \((2, 13)\)

#### Step 1: Identify the center
The center of the circle is given as \((h, k) = (2, -3)\).

#### Step 2: Determine the radius
The point \((2, 13)\) lies on the circle. The radius \(r\) is the distance from the center \((2, -3)\) to the point \((2, 13)\). Using the distance formula:
\[
r = \sqrt{(2 - 2)^2 + (13 - (-3))^2} = \sqrt{0 + (13 + 3)^2} = \sqrt{16^2} = 16
\]

#### Step 3: Write the equation
Using the center \((2, -3)\) and radius \(r = 16\), the equation of the circle is:
\[
(x - 2)^2 + (y + 3)^2 = 16^2
\]
\[
(x - 2)^2 + (y + 3)^2 = 256
\]

#### Final Answer:
\[
\boxed{(x - 2)^2 + (y + 3)^2 = 256}
\]

---

4) Center: \((5, 11)\), Area: \(64\pi\)

#### Step 1: Identify the center
The center of the circle is given as \((h, k) = (5, 11)\).

#### Step 2: Determine the radius
The area of a circle is given by:
\[
A = \pi r^2
\]
We are given \(A = 64\pi\). Solving for \(r\):
\[
64\pi = \pi r^2 \implies r^2 = 64 \implies r = \sqrt{64} = 8
\]

#### Step 3: Write the equation
Using the center \((5, 11)\) and radius \(r = 8\), the equation of the circle is:
\[
(x - 5)^2 + (y - 11)^2 = 8^2
\]
\[
(x - 5)^2 + (y - 11)^2 = 64
\]

#### Final Answer:
\[
\boxed{(x - 5)^2 + (y - 11)^2 = 64}
\]

---

5) Ends of a Diameter: \((-5, 4)\) and \((-5, 18)\)

#### Step 1: Find the center
The center of the circle is the midpoint of the diameter. The midpoint formula is:
\[
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]
Substituting the endpoints \((-5, 4)\) and \((-5, 18)\):
\[
\left( \frac{-5 + (-5)}{2}, \frac{4 + 18}{2} \right) = \left( \frac{-10}{2}, \frac{22}{2} \right) = (-5, 11)
\]
So, the center is \((h, k) = (-5, 11)\).

#### Step 2: Determine the radius
The radius is half the length of the diameter. The length of the diameter is the vertical distance between the points \((-5, 4)\) and \((-5, 18)\):
\[
\text{Diameter} = |18 - 4| = 14
\]
Thus, the radius is:
\[
r = \frac{14}{2} = 7
\]

#### Step 3: Write the equation
Using the center \((-5, 11)\) and radius \(r = 7\), the equation of the circle is:
\[
(x + 5)^2 + (y - 11)^2 = 7^2
\]
\[
(x + 5)^2 + (y - 11)^2 = 49
\]

#### Final Answer:
\[
\boxed{(x + 5)^2 + (y - 11)^2 = 49}
\]

---

6) \(x^2 + y^2 - 8x - 24y + 144 = 0\)

#### Step 1: Rewrite in standard form
Complete the square for \(x\) and \(y\).

- For \(x\): \(x^2 - 8x\)
\[
x^2 - 8x = (x - 4)^2 - 16
\]

- For \(y\): \(y^2 - 24y\)
\[
y^2 - 24y = (y - 12)^2 - 144
\]

Substitute these into the equation:
\[
(x - 4)^2 - 16 + (y - 12)^2 - 144 + 144 = 0
\]
Simplify:
\[
(x - 4)^2 + (y - 12)^2 - 16 = 0
\]
\[
(x - 4)^2 + (y - 12)^2 = 16
\]

#### Final Answer:
\[
\boxed{(x - 4)^2 + (y - 12)^2 = 16}
\]

---

7) Center: \((-3, 12)\), Radius: 1

#### Step 1: Identify the center and radius
The center is \((h, k) = (-3, 12)\) and the radius is \(r = 1\).

#### Step 2: Write the equation
Using the center \((-3, 12)\) and radius \(r = 1\), the equation of the circle is:
\[
(x + 3)^2 + (y - 12)^2 = 1^2
\]
\[
(x + 3)^2 + (y - 12)^2 = 1
\]

#### Final Answer:
\[
\boxed{(x + 3)^2 + (y - 12)^2 = 1}
\]

---

8) \(x^2 + y^2 + 4x - 6y + 12 = 0\), Translated: 1 left and 5 down

#### Step 1: Rewrite in standard form
Complete the square for \(x\) and \(y\).

- For \(x\): \(x^2 + 4x\)
\[
x^2 + 4x = (x + 2)^2 - 4
\]

- For \(y\): \(y^2 - 6y\)
\[
y^2 - 6y = (y - 3)^2 - 9
\]

Substitute these into the equation:
\[
(x + 2)^2 - 4 + (y - 3)^2 - 9 + 12 = 0
\]
Simplify:
\[
(x + 2)^2 + (y - 3)^2 - 1 = 0
\]
\[
(x + 2)^2 + (y - 3)^2 = 1
\]

#### Step 2: Apply the translation
The original center is \((-2, 3)\). Translating 1 unit left and 5 units down:
\[
h' = -2 - 1 = -3, \quad k' = 3 - 5 = -2
\]
The new center is \((-3, -2)\).

#### Step 3: Write the new equation
Using the new center \((-3, -2)\) and radius \(r = 1\), the equation of the circle is:
\[
(x + 3)^2 + (y + 2)^2 = 1
\]

#### Final Answer:
\[
\boxed{(x + 3)^2 + (y + 2)^2 = 1}
\]

---

9) Three Points on the Circle Are: \((-3, -1)\), \((3, 5)\), and \((9, -1)\)

#### Step 1: General equation of a circle
The general form of a circle's equation is:
\[
x^2 + y^2 + Dx + Ey + F = 0
\]

#### Step 2: Substitute the points into the equation
1. For \((-3, -1)\):
\[
(-3)^2 + (-1)^2 + D(-3) + E(-1) + F = 0 \implies 9 + 1 - 3D - E + F = 0 \implies -3D - E + F = -10 \quad \text{(Equation 1)}
\]

2. For \((3, 5)\):
\[
(3)^2 + (5)^2 + D(3) + E(5) + F = 0 \implies 9 + 25 + 3D + 5E + F = 0 \implies 3D + 5E + F = -34 \quad \text{(Equation 2)}
\]

3. For \((9, -1)\):
\[
(9)^2 + (-1)^2 + D(9) + E(-1) + F = 0 \implies 81 + 1 + 9D - E + F = 0 \implies 9D - E + F = -82 \quad \text{(Equation 3)}
\]

#### Step 3: Solve the system of equations
From Equation 1:
\[
-3D - E + F = -10 \quad \text{(Equation 1)}
\]

From Equation 2:
\[
3D + 5E + F = -34 \quad \text{(Equation 2)}
\]

From Equation 3:
\[
9D - E + F = -82 \quad \text{(Equation 3)}
\]

Subtract Equation 1 from Equation 2:
\[
(3D + 5E + F) - (-3D - E + F) = -34 - (-10)
\]
\[
6D + 6E = -24 \implies D + E = -4 \quad \text{(Equation 4)}
\]

Subtract Equation 1 from Equation 3:
\[
(9D - E + F) - (-3D - E + F) = -82 - (-10)
\]
\[
12D = -72 \implies D = -6
\]

Substitute \(D = -6\) into Equation 4:
\[
-6 + E = -4 \implies E = 2
\]

Substitute \(D = -6\) and \(E = 2\) into Equation 1:
\[
-3(-6) - 2 + F = -10
\]
\[
18 - 2 + F = -10 \implies 16 + F = -10 \implies F = -26
\]

#### Step 4: Write the equation
The equation is:
\[
x^2 + y^2 - 6x + 2y - 26 = 0
\]

#### Step 5: Convert to standard form
Complete the square for \(x\) and \(y\).

- For \(x\): \(x^2 - 6x\)
\[
x^2 - 6x = (x - 3)^2 - 9
\]

- For \(y\): \(y^2 + 2y\)
\[
y^2 + 2y = (y + 1)^2 - 1
\]

Substitute these into the equation:
\[
(x - 3)^2 - 9 + (y + 1)^2 - 1 - 26 = 0
\]
Simplify:
\[
(x - 3)^2 + (y + 1)^2 - 36 = 0
\]
\[
(x - 3)^2 + (y + 1)^2 = 36
\]

#### Final Answer:
\[
\boxed{(x - 3)^2 + (y + 1)^2 = 36}
\]

---

10) \((x - 8)^2 + (y - 7)^2 = 25\), Translated: 3 left and 4 up

#### Step 1: Identify the original center and radius
The original center is \((8, 7)\) and the radius is \(r = 5\).

#### Step 2: Apply the translation
Translating 3 units left and 4 units up:
\[
h' = 8 - 3 = 5, \quad k' = 7 + 4 = 11
\]
The new center is \((5, 11)\).

#### Step 3: Write the new equation
Using the new center \((5, 11)\) and radius \(r = 5\), the equation of the circle is:
\[
(x - 5)^2 + (y - 11)^2 = 25
\]

#### Final Answer:
\[
\boxed{(x - 5)^2 + (y - 11)^2 = 25}
\]

---

Final Answers:

1. \(\boxed{(x - 9)^2 + (y + 8)^2 = 100}\)
2. \(\boxed{(x - 1)^2 + (y - 9)^2 = 36}\)
3. \(\boxed{(x - 2)^2 + (y + 3)^2 = 256}\)
4. \(\boxed{(x - 5)^2 + (y - 11)^2 = 64}\)
5. \(\boxed{(x + 5)^2 + (y - 11)^2 = 49}\)
6. \(\boxed{(x - 4)^2 + (y - 12)^2 = 16}\)
7. \(\boxed{(x + 3)^2 + (y - 12)^2 = 1}\)
8. \(\boxed{(x + 3)^2 + (y + 2)^2 = 1}\)
9. \(\boxed{(x - 3)^2 + (y + 1)^2 = 36}\)
10. \(\boxed{(x - 5)^2 + (y - 11)^2 = 25}\)