Equations of Circles

Write and graph equations of circles

Equations of circles content

📚 Practice Problems

1Problem 1easy

Question:

Write the equation of a circle with center (3, -2) and radius 5.

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Step 1: Recall the standard form of a circle: (x - h)² + (y - k)² = r² where (h, k) is the center and r is the radius

Step 2: Identify the values: Center: (h, k) = (3, -2) Radius: r = 5

Step 3: Substitute into the formula: (x - 3)² + (y - (-2))² = 5² (x - 3)² + (y + 2)² = 25

Answer: (x - 3)² + (y + 2)² = 25

2Problem 2easy

Question:

Find the center and radius of the circle: (x + 4)² + (y - 1)² = 36

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Step 1: Recall standard form: (x - h)² + (y - k)² = r²

Step 2: Rewrite the equation to match standard form: (x + 4)² + (y - 1)² = 36 (x - (-4))² + (y - 1)² = 6²

Step 3: Identify h, k, and r: h = -4 (note: x + 4 = x - (-4)) k = 1 r² = 36, so r = 6

Step 4: State the center and radius: Center: (-4, 1) Radius: 6

Answer: Center (-4, 1), radius 6

3Problem 3medium

Question:

Find the equation of a circle with center (2, 5) that passes through point (6, 8).

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Step 1: Find the radius using distance formula: The radius is the distance from center to the point r = √[(6 - 2)² + (8 - 5)²] r = √[4² + 3²] r = √[16 + 9] r = √25 r = 5

Step 2: Write the equation: (x - 2)² + (y - 5)² = 5² (x - 2)² + (y - 5)² = 25

Step 3: Verify the point is on the circle: (6 - 2)² + (8 - 5)² = 4² + 3² = 16 + 9 = 25 ✓

Answer: (x - 2)² + (y - 5)² = 25

4Problem 4medium

Question:

Convert the equation x² + y² - 6x + 4y - 12 = 0 to standard form and find the center and radius.

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Step 1: Group x terms and y terms: (x² - 6x) + (y² + 4y) = 12

Step 2: Complete the square for x: x² - 6x → add (6/2)² = 9 (x² - 6x + 9) = (x - 3)²

Step 3: Complete the square for y: y² + 4y → add (4/2)² = 4 (y² + 4y + 4) = (y + 2)²

Step 4: Add the same values to the right side: (x² - 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4 (x - 3)² + (y + 2)² = 25

Step 5: Identify center and radius: Center: (3, -2) Radius: √25 = 5

Answer: Standard form: (x - 3)² + (y + 2)² = 25 Center: (3, -2), Radius: 5

5Problem 5hard

Question:

A circle passes through points A(1, 2), B(5, 4), and C(3, 6). Find the equation of the circle.

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Step 1: Use general form: x² + y² + Dx + Ey + F = 0

Step 2: Substitute point A(1, 2): 1² + 2² + D(1) + E(2) + F = 0 1 + 4 + D + 2E + F = 0 D + 2E + F = -5 ... equation (1)

Step 3: Substitute point B(5, 4): 5² + 4² + D(5) + E(4) + F = 0 25 + 16 + 5D + 4E + F = 0 5D + 4E + F = -41 ... equation (2)

Step 4: Substitute point C(3, 6): 3² + 6² + D(3) + E(6) + F = 0 9 + 36 + 3D + 6E + F = 0 3D + 6E + F = -45 ... equation (3)

Step 5: Solve the system (subtract equations): From (2) - (1): 4D + 2E = -36 → 2D + E = -18 ... (4) From (3) - (1): 2D + 4E = -40 → D + 2E = -20 ... (5)

Step 6: Solve equations (4) and (5): From (4): E = -18 - 2D Substitute into (5): D + 2(-18 - 2D) = -20 D - 36 - 4D = -20 -3D = 16 D = -16/3

Then: E = -18 - 2(-16/3) = -18 + 32/3 = -54/3 + 32/3 = -22/3

Step 7: Find F from equation (1): D + 2E + F = -5 -16/3 + 2(-22/3) + F = -5 -16/3 - 44/3 + F = -5 -60/3 + F = -5 -20 + F = -5 F = 15

Step 8: Write the equation: x² + y² - (16/3)x - (22/3)y + 15 = 0

Multiply by 3: 3x² + 3y² - 16x - 22y + 45 = 0

Step 9: Convert to standard form (complete the square): 3(x² - 16x/3) + 3(y² - 22y/3) = -45 x² - (16/3)x + y² - (22/3)y = -15

Complete squares: (x - 8/3)² + (y - 11/3)² = (8/3)² + (11/3)² - 15 = 64/9 + 121/9 - 135/9 = 50/9

Center: (8/3, 11/3) Radius: √(50/9) = 5√2/3

Answer: 3x² + 3y² - 16x - 22y + 45 = 0 Or: (x - 8/3)² + (y - 11/3)² = 50/9