Equations of Circles Worksheet - Kuta Software - Free Printable

Let's solve each problem step by step to write the standard form equation of a circle:

The standard form of a circle is:
$$
(x - h)^2 + (y - k)^2 = r^2
$$
where $(h, k)$ is the center and $r$ is the radius.

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1) $8x + x^2 - 2y = 64 - y^2$

Rearrange all terms to one side:
$$
x^2 + 8x + y^2 - 2y - 64 = 0
$$

Group $x$ and $y$ terms:
$$
(x^2 + 8x) + (y^2 - 2y) = 64
$$

Complete the square:

- $x^2 + 8x$: take half of 8 → 4, square it → 16
- $y^2 - 2y$: half of -2 → -1, square it → 1

Add these to both sides:
$$
(x^2 + 8x + 16) + (y^2 - 2y + 1) = 64 + 16 + 1 = 81
$$

Factor:
$$
(x + 4)^2 + (y - 1)^2 = 81
$$

✔ Answer: $(x + 4)^2 + (y - 1)^2 = 81$

---

2) $137 + 6y = -y^2 - 24x$

Bring all terms to one side:
$$
y^2 + 6y + 24x + 137 = 0
$$

Wait — we need both $x^2$ and $y^2$. But here only $y^2$ appears. Let’s check again.

Original:
$137 + 6y = -y^2 - 24x$

Move all to left:
$$
137 + 6y + y^2 + 24x = 0 \Rightarrow y^2 + 6y + 24x + 137 = 0
$$

Still no $x^2$. This suggests something is wrong — unless there's a typo?

But wait: perhaps the original equation has an $x^2$ term missing? Let's recheck.

Wait — maybe it's supposed to be symmetric? Let's assume the user meant this equation includes $x^2$, but it's not written. Alternatively, let's suppose it's a typo and should have $x^2$. But as given, it doesn't have $x^2$.

Alternatively, maybe it's:
$$
137 + 6y = -y^2 - 24x \quad \text{and} \quad x^2 \text{ is missing}
$$

This can't represent a circle without $x^2$. So likely a typo.

But let's suppose it's actually:
$$
137 + 6y = -y^2 - 24x - x^2
\Rightarrow x^2 + 24x + y^2 + 6y = -137
$$

Now that makes sense.

So:
$$
x^2 + 24x + y^2 + 6y = -137
$$

Complete the square:

- $x^2 + 24x$: half of 24 → 12, square → 144
- $y^2 + 6y$: half of 6 → 3, square → 9

Add to both sides:
$$
x^2 + 24x + 144 + y^2 + 6y + 9 = -137 + 144 + 9 = 16
$$

Factor:
$$
(x + 12)^2 + (y + 3)^2 = 16
$$

✔ Answer: $(x + 12)^2 + (y + 3)^2 = 16$

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3) $x^2 + y^2 + 14x - 12y + 4 = 0$

Group:
$$
(x^2 + 14x) + (y^2 - 12y) = -4
$$

Complete the square:

- $x^2 + 14x$: half of 14 → 7, square → 49
- $y^2 - 12y$: half of -12 → -6, square → 36

Add to both sides:
$$
(x^2 + 14x + 49) + (y^2 - 12y + 36) = -4 + 49 + 36 = 81
$$

Factor:
$$
(x + 7)^2 + (y - 6)^2 = 81
$$

✔ Answer: $(x + 7)^2 + (y - 6)^2 = 81$

---

4) $y^2 + 2x + x^2 = 24y - 120$

Rewriting:
$$
x^2 + y^2 + 2x - 24y = -120
$$

Group:
$$
(x^2 + 2x) + (y^2 - 24y) = -120
$$

Complete the square:

- $x^2 + 2x$: half of 2 → 1, square → 1
- $y^2 - 24y$: half of -24 → -12, square → 144

Add:
$$
(x^2 + 2x + 1) + (y^2 - 24y + 144) = -120 + 1 + 144 = 25
$$

Factor:
$$
(x + 1)^2 + (y - 12)^2 = 25
$$

✔ Answer: $(x + 1)^2 + (y - 12)^2 = 25$

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5) $x^2 + 2x + y^2 = 55 + 10y$

Bring all to left:
$$
x^2 + 2x + y^2 - 10y - 55 = 0
$$

Group:
$$
(x^2 + 2x) + (y^2 - 10y) = 55
$$

Complete the square:

- $x^2 + 2x$: add 1 → $(x+1)^2$
- $y^2 - 10y$: add 25 → $(y-5)^2$

Add to both sides:
$$
(x+1)^2 + (y-5)^2 = 55 + 1 + 25 = 81
$$

✔ Answer: $(x + 1)^2 + (y - 5)^2 = 81$

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6) $8x + 32y + y^2 = -263 - x^2$

Bring all to left:
$$
x^2 + 8x + y^2 + 32y + 263 = 0
$$

Group:
$$
(x^2 + 8x) + (y^2 + 32y) = -263
$$

Complete the square:

- $x^2 + 8x$: half of 8 → 4, square → 16
- $y^2 + 32y$: half of 32 → 16, square → 256

Add:
$$
(x^2 + 8x + 16) + (y^2 + 32y + 256) = -263 + 16 + 256 = -1
$$

Wait: $-263 + 16 + 256 = -1$

So:
$$
(x + 4)^2 + (y + 16)^2 = -1
$$

✘ Not possible — radius squared is negative.

Check calculation:

- $-263 + 16 = -247$
- $-247 + 256 = 9$

Ah! I made a mistake.

So:
$$
(x+4)^2 + (y+16)^2 = -263 + 16 + 256 = 9
$$

✔ Corrected: $(x + 4)^2 + (y + 16)^2 = 9$

✔ Answer: $(x + 4)^2 + (y + 16)^2 = 9$

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7) Center: $(-11, -8)$, Radius: 4

Plug into standard form:
$$
(x + 11)^2 + (y + 8)^2 = 4^2 = 16
$$

✔ Answer: $(x + 11)^2 + (y + 8)^2 = 16$

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8) Center: $(-6, -1.5)$, Radius: $\sqrt{5}$

So:
$$
(x + 6)^2 + \left(y + \frac{3}{2}\right)^2 = (\sqrt{5})^2 = 5
$$

✔ Answer: $(x + 6)^2 + \left(y + \frac{3}{2}\right)^2 = 5$

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9) $(x - 16)^2 + (y - 6)^2 = 1$, translated 4 left, 2 up

Translation: subtract 4 from $x$, add 2 to $y$

New center: $(16 - 4, 6 + 2) = (12, 8)$

So new equation:
$$
(x - 12)^2 + (y - 8)^2 = 1
$$

✔ Answer: $(x - 12)^2 + (y - 8)^2 = 1$

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10) $(x + 5)^2 + (y + 7)^2 = 36$, translated 5 left, 4 down

Current center: $(-5, -7)$

Translate: $x$: $-5 - 5 = -10$, $y$: $-7 - 4 = -11$

New center: $(-10, -11)$

Equation:
$$
(x + 10)^2 + (y + 11)^2 = 36
$$

✔ Answer: $(x + 10)^2 + (y + 11)^2 = 36$

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11) Graph: Circle centered at origin with radius 5

From graph: center at $(0, 0)$, goes from $-5$ to $5$ on both axes → radius = 5

So:
$$
x^2 + y^2 = 25
$$

✔ Answer: $x^2 + y^2 = 25$

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12) Graph: Small circle in fourth quadrant

From grid:

- Center appears at $(3, -2)$
- Radius: from center to edge — about 2 units

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

✔ Answer: $(x - 3)^2 + (y + 2)^2 = 4$

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✔ Final Answers:

1) $(x + 4)^2 + (y - 1)^2 = 81$
2) $(x + 12)^2 + (y + 3)^2 = 16$
3) $(x + 7)^2 + (y - 6)^2 = 81$
4) $(x + 1)^2 + (y - 12)^2 = 25$
5) $(x + 1)^2 + (y - 5)^2 = 81$
6) $(x + 4)^2 + (y + 16)^2 = 9$
7) $(x + 11)^2 + (y + 8)^2 = 16$
8) $(x + 6)^2 + \left(y + \frac{3}{2}\right)^2 = 5$
9) $(x - 12)^2 + (y - 8)^2 = 1$
10) $(x + 10)^2 + (y + 11)^2 = 36$
11) $x^2 + y^2 = 25$
12) $(x - 3)^2 + (y + 2)^2 = 4$

Let me know if you'd like explanations for any specific one!