Writing Equations of Circles Worksheet for 10th - 12th Grade ... - Free Printable

It looks like you've uploaded a worksheet titled "Writing Equations of Circles" with several problems asking to write the standard form equation of a circle given certain information (center and radius, or graph). Let's go through each problem step by step and solve them.

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🔷 Standard Form of a Circle:

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,
- $r$ is the radius.

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## ✔ Problem 1: Center at $(-3, 4)$, Radius = 5

We plug into the formula:

$$
(x - (-3))^2 + (y - 4)^2 = 5^2
\Rightarrow (x + 3)^2 + (y - 4)^2 = 25
$$

✔ Answer:
$$
(x + 3)^2 + (y - 4)^2 = 25
$$

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## ✔ Problem 2: Center at $(0, -2)$, Radius = 7

$$
(x - 0)^2 + (y - (-2))^2 = 7^2
\Rightarrow x^2 + (y + 2)^2 = 49
$$

✔ Answer:
$$
x^2 + (y + 2)^2 = 49
$$

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## ✔ Problem 3: Center at $(-6, -2)$, Radius = 3

$$
(x + 6)^2 + (y + 2)^2 = 3^2
\Rightarrow (x + 6)^2 + (y + 2)^2 = 9
$$

✔ Answer:
$$
(x + 6)^2 + (y + 2)^2 = 9
$$

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## ✔ Problem 4: Center at $(-4, -1)$, Radius = 1.5

Note: $1.5 = \frac{3}{2}$, so $r^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$

$$
(x + 4)^2 + (y + 1)^2 = \frac{9}{4}
$$

✔ Answer:
$$
(x + 4)^2 + (y + 1)^2 = \frac{9}{4}
$$

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## ✔ Problem 5: Graph (a)

Looking at the graph labeled (a), we see:
- The circle is centered at the origin $(0, 0)$,
- It passes through $(0, 3)$ and $(3, 0)$, so the radius is 3.

So:
$$
(x - 0)^2 + (y - 0)^2 = 3^2
\Rightarrow x^2 + y^2 = 9
$$

✔ Answer:
$$
x^2 + y^2 = 9
$$

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## ✔ Problem 6: Graph (b)

From the graph labeled (b):
- The center appears to be at $(3, -1)$,
- The circle goes from $x = 1$ to $x = 5$, so diameter = 4 → radius = 2.
- Similarly, from $y = -3$ to $y = 1$, so radius = 2.

So:
- Center: $(3, -1)$
- Radius: $2$

$$
(x - 3)^2 + (y + 1)^2 = 4
$$

✔ Answer:
$$
(x - 3)^2 + (y + 1)^2 = 4
$$

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📌 Summary of Answers:

| Problem | Equation |
|--------|---------|
| 1 | $(x + 3)^2 + (y - 4)^2 = 25$ |
| 2 | $x^2 + (y + 2)^2 = 49$ |
| 3 | $(x + 6)^2 + (y + 2)^2 = 9$ |
| 4 | $(x + 4)^2 + (y + 1)^2 = \frac{9}{4}$ |
| 5 (Graph a) | $x^2 + y^2 = 9$ |
| 6 (Graph b) | $(x - 3)^2 + (y + 1)^2 = 4$ |

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Let me know if you'd like these rewritten in general form or explained further!