Different Forms of Equations of a Circle

1

easyFind the equation of a circle with center (0,0) and radius 3.

A. x² + y² = 3 B. x² + y² = 9 C. x² + y² = 6 D. x² + y² = 0

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✔ Correct: B

Using x² + y² = r², with r = 3, we get x² + y² = 3² = 9.

2

easyWhat is the radius of the circle x² + y² = 49?

A. 7 B. 49 C. 14 D. 24.5

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✔ Correct: A

Comparing x² + y² = r² with x² + y² = 49, r² = 49, so r = √49 = 7.

3

mediumFind the center of the circle (x + 5)² + (y − 2)² = 16.

A. (5, -2) B. (-5, 2) C. (5, 2) D. (-5, -2)

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✔ Correct: B

Comparing with (x − h)² + (y − k)² = r², h = -5 and k = 2.

4

mediumIf the center of a circle is (1, 2) and it passes through (4, 6), what is the radius?

A. 3 B. 4 C. 5 D. 6

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✔ Correct: C

Radius = distance between (1,2) and (4,6) = √((4-1)² + (6-2)²) = √(3² + 4²) = √25 = 5.

5

hardFind the center of x² + y² − 8x + 10y + 5 = 0.

A. (4, -5) B. (-4, 5) C. (8, -10) D. (-8, 10)

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✔ Correct: A

2g = -8 → g = -4; 2f = 10 → f = 5. Center is (-g, -f) = (4, -5).

6

hardFind the radius of x² + y² + 2x − 4y − 4 = 0.

A. 1 B. 2 C. 3 D. 4

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✔ Correct: C

g = 1, f = -2, c = -4. Radius = √(g² + f² - c) = √(1² + (-2)² - (-4)) = √(1 + 4 + 4) = √9 = 3.

7

very hardA circle has center (3, 4) and touches the y-axis. What is its equation?

A. (x-3)² + (y-4)² = 9 B. (x-3)² + (y-4)² = 16 C. (x-3)² + (y-4)² = 25 D. (x-3)² + (y-4)² = 7

Show Answer

✔ Correct: B

If it touches the y-axis, radius = |h| = |3| = 3. Wait, if it touches y-axis, radius = |h| = 3. Equation is (x-3)² + (y-4)² = 9.

1

Assertion (A): The circle x² + y² = 25 has center (0,0) and radius 5.
Reason (R): The standard form x�� + y² = r² represents a circle centered at the origin with radius r.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: A

Both assertion and reason are true, and the reason correctly explains the assertion.

2

Assertion (A): The circle x² + y² + 4x + 6y + 13 = 0 has radius 0.
Reason (R): The radius formula is √(g² + f² - c). Here g=2, f=3, c=13, so √(4+9-13) = 0.

Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is not the correct explanation of A.
(C) A is true but R is false.
(D) A is false but R is true.

A. Option A B. Option B C. Option C D. Option D

Show Answer

✔ Correct: A

Both are true. A circle with radius 0 is a point circle.

1

mediumGiven the circle (x - 1)² + (y + 2)² = 25, find: (i) Center (ii) Radius (iii) x-intercepts.

Show Answer & Solution

(i) Center is (1, -2).
(ii) Radius is √25 = 5.
(iii) Set y=0: (x-1)² + 4 = 25 → (x-1)² = 21 → x = 1 ± √21.

2

mediumGiven the circle x² + y² - 6x - 8y = 0, find: (i) Center (ii) Radius (iii) Does it pass through (0,0)?

Show Answer & Solution

(i) 2g = -6, 2f = -8 → Center (3, 4).
(ii) Radius = √(3² + 4² - 0) = 5.
(iii) Substitute (0,0): 0² + 0² - 6(0) - 8(0) = 0. Yes, it passes through (0,0).

3

hardGiven a circle with center (2, 3) passing through (5, 7), find: (i) Radius (ii) Equation (iii) y-intercepts.

Show Answer & Solution

(i) r = √((5-2)² + (7-3)²) = √(3² + 4²) = 5.
(ii) (x-2)² + (y-3)² = 25.
(iii) Set x=0: (-2)² + (y-3)² = 25 → (y-3)² = 21 → y = 3 ± √21.

1

hardFind the equation of the circle passing through (0,0), (2,0), and (0,2).

Show Answer & Solution

Step 1: Let equation be x² + y² + 2gx + 2fy + c = 0.
Step 2: Pass through (0,0) → c = 0.
Step 3: Pass through (2,0) → 4 + 0 + 4g + 0 + 0 = 0 → g = -1.
Step 4: Pass through (0,2) → 0 + 4 + 0 + 4f + 0 = 0 → f = -1.
Step 5: Equation is x² + y² - 2x - 2y = 0.

2

very hardConvert x² + y² + 10x - 6y + 9 = 0 into center-radius form and identify properties.

Show Answer & Solution

Step 1: Group terms: (x² + 10x) + (y² - 6y) = -9.
Step 2: Complete squares: (x² + 10x + 25) + (y² - 6y + 9) = -9 + 25 + 9.
Step 3: Simplify: (x + 5)² + (y - 3)² = 25.
Step 4: Identify center: (-5, 3).
Step 5: Identify radius: √25 = 5.

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Case Study: A circular park is designed on a coordinate grid where each unit represents 10 meters. The park's boundary is defined by the equation x² + y² - 8x - 6y = 0.

Sub-question 1: (i) Find the center of the park in coordinate units. (ii) Calculate the radius in meters.

Show Solution

(i) 2g = -8, 2f = -6 → Center (4, 3).
(ii) r = √(4² + 3²) = 5 units. Since 1 unit = 10m, radius = 5 × 10 = 50 meters.

Sub-question 2: If a fountain is placed at the center and a path is built along the boundary, find the total length of the boundary (circumference).

Show Solution

Circumference = 2πr. r = 50m. Circumference = 2 × π × 50 = 100π meters.

Standard Form of a Circle

Center-Radius Form

General Form of a Circle

Section A: Multiple Choice Questions (MCQs)

Section B: Assertion–Reason Questions

Section C: Short Answer Questions

Section D: Long Answer Questions

Section E: Case-Based Question

CBSE Board & Sample Paper Past Year Questions

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