Equation of a Circle Worksheets - Math Monks - Free Printable

Let's solve each problem step-by-step from the "Equation of a Circle Worksheet".

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🔷 General Formula for the Equation of a Circle:

The standard form of the equation of a circle with center $(h, k)$ and radius $r$ is:

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

For a circle centered at the origin $(0, 0)$, this simplifies to:

$$
x^2 + y^2 = r^2
$$

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## ✔ Problem 1:
Which of the following is the equation of a circle with center at (0,0) and a radius of 6 units?

- Radius $r = 6$, so $r^2 = 36$
- Equation: $x^2 + y^2 = 36$

✔ Answer: (c) $x^2 + y^2 = 36$

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## ✔ Problem 2:
Which of the following is the equation of a circle with center at (0,0) and a radius of 74 units?

- $r = 74$, so $r^2 = 74^2 = 5476$
- Equation: $x^2 + y^2 = 5476$

✔ Answer: (a) $x^2 + y^2 = 5476$

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## ✔ Problem 3:
Which of the following is the equation of a circle with center at (0,0) and a diameter of 28 units?

- Diameter = 28 → Radius = $ \frac{28}{2} = 14 $
- $r^2 = 14^2 = 196$
- Equation: $x^2 + y^2 = 196$

✔ Answer: (d) $x^2 + y^2 = 196$

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## ✔ Problem 4:
Which of the following is the equation of a circle with center at (0,0) and a diameter of 28 units?

This is identical to Problem 3, so same answer.

✔ Answer: (d) $x^2 + y^2 = 196$

> Note: There might be a typo in the worksheet — Problems 3 and 4 are the same. But both should have answer (d).

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## ✔ Problem 5:
Center = (2, -4), Radius = 15

Use standard form:
$$
(x - h)^2 + (y - k)^2 = r^2
$$
Plug in $h = 2$, $k = -4$, $r = 15$, $r^2 = 225$:

$$
(x - 2)^2 + (y + 4)^2 = 225
$$

✔ Answer: $(x - 2)^2 + (y + 4)^2 = 225$

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## ✔ Problem 6:
Center = (5, 1), and a point on the circle (8, -2)

We know center $(h,k) = (5,1)$, and a point $(x,y) = (8,-2)$ lies on the circle.

Find radius using distance formula:

$$
r = \sqrt{(8 - 5)^2 + (-2 - 1)^2} = \sqrt{(3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18}
$$

So $r^2 = 18$

Now write equation:

$$
(x - 5)^2 + (y - 1)^2 = 18
$$

✔ Answer: $(x - 5)^2 + (y - 1)^2 = 18$

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## ✔ Problem 7:
Circle with (5,1) and (3,-1) as endpoints of the diameter

To find the equation:

Step 1: Find the center (midpoint of diameter)

$$
\text{Center} = \left( \frac{5+3}{2}, \frac{1 + (-1)}{2} \right) = \left( \frac{8}{2}, \frac{0}{2} \right) = (4, 0)
$$

Step 2: Find radius (half the length of diameter)

Distance between (5,1) and (3,-1):

$$
d = \sqrt{(5-3)^2 + (1 - (-1))^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8}
$$

So radius $r = \frac{\sqrt{8}}{2} = \sqrt{2}$, but we need $r^2$:

$$
r^2 = \left(\frac{\sqrt{8}}{2}\right)^2 = \frac{8}{4} = 2
$$

Alternatively, use half the squared distance:

$$
r^2 = \frac{1}{4} \times ( (5-3)^2 + (1+1)^2 ) = \frac{1}{4}(4 + 4) = \frac{8}{4} = 2
$$

So equation:

$$
(x - 4)^2 + (y - 0)^2 = 2
\Rightarrow (x - 4)^2 + y^2 = 2
$$

✔ Answer: $(x - 4)^2 + y^2 = 2$

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## ✔ Problem 8:
Same as Problem 7?
Yes — it says "Circle with (5,1) and (3,-1) as endpoints of the diameter."

So same answer.

✔ Answer: $(x - 4)^2 + y^2 = 2$

> Possibly a duplicate question.

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## ✔ Problem 9:
Rewrite in standard form: $x^2 + y^2 - 12x + 8y + 32 = 0$

Group terms:

$$
(x^2 - 12x) + (y^2 + 8y) + 32 = 0
$$

Complete the square:

- $x^2 - 12x$: take half of -12 → -6, square = 36
- $y^2 + 8y$: half of 8 → 4, square = 16

Add and subtract these inside the equation:

$$
(x^2 - 12x + 36) + (y^2 + 8y + 16) + 32 - 36 - 16 = 0
$$

$$
(x - 6)^2 + (y + 4)^2 - 20 = 0
\Rightarrow (x - 6)^2 + (y + 4)^2 = 20
$$

✔ Answer: $(x - 6)^2 + (y + 4)^2 = 20$

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## ✔ Problem 10:
Rewrite in standard form: $x^2 + y^2 + 22x - 22y - 20 = 0$

Group:

$$
(x^2 + 22x) + (y^2 - 22y) - 20 = 0
$$

Complete the square:

- $x^2 + 22x$: half of 22 = 11, square = 121
- $y^2 - 22y$: half of -22 = -11, square = 121

Add and subtract:

$$
(x^2 + 22x + 121) + (y^2 - 22y + 121) - 20 - 121 - 121 = 0
$$

$$
(x + 11)^2 + (y - 11)^2 - 262 = 0
\Rightarrow (x + 11)^2 + (y - 11)^2 = 262
$$

✔ Answer: $(x + 11)^2 + (y - 11)^2 = 262$

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📝 Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | (c) $x^2 + y^2 = 36$ |
| 2 | (a) $x^2 + y^2 = 5476$ |
| 3 | (d) $x^2 + y^2 = 196$ |
| 4 | (d) $x^2 + y^2 = 196$ |
| 5 | $(x - 2)^2 + (y + 4)^2 = 225$ |
| 6 | $(x - 5)^2 + (y - 1)^2 = 18$ |
| 7 | $(x - 4)^2 + y^2 = 2$ |
| 8 | $(x - 4)^2 + y^2 = 2$ |
| 9 | $(x - 6)^2 + (y + 4)^2 = 20$ |
| 10 | $(x + 11)^2 + (y - 11)^2 = 262$ |

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