Solved Examples and Worksheet for Equations of Circles

Q1Find the equation of the circle in standard form with center as origin and radius 35.

A. x2 + y2 = 5
B. x2 - y2 = 3
C. x2 + y2 = 45
D. x2 + y2 = 0

Solutions
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Step: 1
The equation of the circle in standard form with center (0, 0) and radius 35 is (x - 0)2 + (y - 0)2 = (35)2
  [Use formula.]
Step: 2
⇒ x2 + y2 = 45
Correct Answer is :   x2 + y2 = 45
Q2Find the center and radius of the circle.
(x - 5)2 + (y - 2)2 = 25
A. (- 5, 2); 5
B. (5, 2); 5
C. (5, 2); 25
D. (5, - 2); 5

Solutions
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Step: 1
(x - 5)2 + (y - 2)2 = 25
Step: 2
Center (h, k) = (5, 2)
  [Compare with standard form of the circle (x - h)2 + (y - k)2 = r2.]
Step: 3
Radius, r = 25 = 5
Correct Answer is :   (5, 2); 5
Q3Write the standard equation of the circle with center (6, - 3) and radius 7.

A. (x + 6)² + (y - 3)² = 49
B. (x - 6)² + (y + 3)² = 49
C. (x - 6)² + (y - 3)² = 49
D. (x + 6)² + (y + 3)² = 49

Solutions
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Step: 1
The standard form of an equation of a circle with center (h, k) and radius r is (x - h)² + (y - k)² = r²
Step: 2
(h, k) = (6, - 3) and r = 7
  [Compare with the standard form.]
Step: 3
(x - 6)² + (y - (- 3))² = 49
  [Substitute.]
Step: 4
(x - 6)² + (y + 3)² = 49
  [Simplify.]
Correct Answer is :   (x - 6)² + (y + 3)² = 49
Q4Find the coordinates of the center of the circle whose equation is (x - 13)² + (y + 11)² = 225.
A. (13, - 11)
B. (-13, 11)
C. (- 13, - 11)
D. (13, 11)

Solutions
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Step: 1
The standard form of an equation of a circle with center (h, k) and radius r is (x - h)² + (y - k)² = r²
Step: 2
(x - 13)² + (y + 11)² = 225
  [Given.]
Step: 3
(x - 13)² + (y - (- 11))² = 15²
Step: 4
(h, k) = (13, - 11)
  [Compare with standard form.]
Step: 5
The coordinates of the center of the required circle is (13, - 11).
Correct Answer is :   (13, - 11)
Q5What is the center of the circle whose equation is x² + y² - 6x - 8y - 6 = 0?

A. (3, 4)
B. (- 3 , 4)
C. (4, 3)
D. (- 3, - 4)

Solutions
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Step: 1
x² + y² - 6x - 8y - 6 = 0
  [Given equation.]
Step: 2
(x² - 6x) + (y² - 8y) = 6
  [Group the terms.]
Step: 3
(x² - 6x + 9) + (y² - 8y + 16) = 31
  [Add 9, 16 on both sides.]
Step: 4
(x - 3)² + (y - 4)² = 31
Step: 5
Center = (3, 4)
  [Compare with standard form.]
Correct Answer is :   (3, 4)
Q6What is the radius of the circle whose equation is x² + y² - 6x + 4y = 12?

A. 7 units
B. 5 units
C. 9 units
D. 8 units

Solutions
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Step: 1
x² + y² - 6x + 4y = 12
  [Given equation.]
Step: 2
(x² - 6x) + (y² + 4y) = 12
  [Group the terms.]
Step: 3
(x² - 6x + 9) + (y² + 4y + 4) = 12 + 13
  [Add 9, 4 on both sides.]
Step: 4
(x - 3)² + (y + 2)² = 25
Step: 5
(x - 3)² + (y + 2)² = 5²
  [Put in standard form.]
Step: 6
Radius = r = 5 units
  [Compare with the standard form.]
Correct Answer is :   5 units
Q7What is the equation of the circle with center (5, - 3) and passing through (2, - 2)?

A. (x + 5)² + (y + 3)² = 81
B. (x - 5)² + (y - 3)² = 10
C. (x - 5)² + (y + 3)² = 100
D. (x - 5)² + (y + 3)² = 10

Solutions
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Step: 1
The radius is the distance from the center to any point on the circle.
Step: 2
center = (5, - 3) and (2, - 2) lies on the circle
  [Given.]
Step: 3
radius = r = (5 - 2)² + (-3 - (-2))²
  [Use the distance formula.]
Step: 4
r = 9 + 1 = 10
  [Simplify.]
Step: 5
The standard form of an equation of a circle with center (h, k) and radius r is (x - h)² + (y - k)² = r²
Step: 6
(h , k) = (5, - 3) and r = 10
  [Compare with the standard form.]
Step: 7
(x - 5)² + (y - (- 3))² = (10)²
  [Substitute.]
Step: 8
(x - 5)² + (y + 3)² = 10
  [Simplify.]
Correct Answer is :   (x - 5)² + (y + 3)² = 10
Q8If P(- 18, 0), Q(18, 0) are the endpoints of the diameter of a circle, then what is the equation of the circle?

A. x2 + y2 = 0
B. x2 + y2 = 324
C. x2 + y2 = 18
D. x2 + y2 = 36

Solutions
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Step: 1
PQ = (18 - (- 18))2+(0 - 0)2
  [Use the distance formula (x2-x1)2+(y2-y1)2.]
Step: 2
= (36)2 + 0 = 36
  
Step: 3
Diameter = 36 ⇒ radius = 362 = 18
Step: 4
Center of the circle (- 18 + 182, 0 + 02) = (0, 0)
Step: 5
Equation of a circle with center (0, 0) and radius 18 is
(x - 0)2 + (y - 0)2 = 182
  [Use (x - h)2 + (y - k)2 = r2.]
Step: 6
x2 + y2 = 324
Correct Answer is :   x2 + y2 = 324
Q9Find the center and radius of the circle (x - 6)2 + y2 = 4.

A. (0, 6), 2
B. (6, 0), 2
C. (6, 0), 4
D. (- 6, 0), 2

Solutions
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Step: 1
(x - 6)2 + y2 = 4
  [Equation of the circle.]
Step: 2
(h, k) = (6, 0) and r = 2
  [Compare with (x - h)2 + (y - k)2 = r2.]
Step: 3
So, the center of the circle is (6, 0) and its radius is 2 units.
Correct Answer is :   (6, 0), 2
Q10Find the center and radius of the circle x2 + y2 - 8ax - 14ay + 16a2 = 0.

A. (4a, 7a), 7a
B. (4a, 7a), 49a2
C. (7a, - 4a), 7a
D. (4a, 7a), 49a

Solutions
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Step: 1
x2 + y2 - 8ax - 14ay + 16a2 = 0
Step: 2
(x2 - 8ax + ___ ) + (y2 - 14ay + ___ ) = - 16a2
  [Group the terms.]
Step: 3
(x2 - 8ax + 16a2) + (y2 - 14ay + 49a2) = - 16a2 + 16a2 + 49a2
Step: 4
(x - 4a)2 + (y - 7a)2 = (7a)2
  [Complete the squares.]
Step: 5
(h, k) = (4a, 7a) and r = 7a
  [Compare with (x - h)2 + (y - k)2 = (r)2.]
Step: 6
So, the center of the circle is (4a, 7a) and radius is 7a.
Correct Answer is :   (4a, 7a), 7a
Q11What is the equation of a degenerate circle with center (- 3, 5)?

A. (x + 3)2 + (y - 5)2 = 34
B. (x - 3)2 + (y + 5)2 = 8
C. (x - 3)2 + (y + 5)2 = 0
D. (x + 3)2 + (y - 5)2 = 0

Solutions
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Step: 1
The radius of a degenerate circle is zero.
Step: 2
The equation of a degenerate circle is (x + 3)2 + (y - 5)2 = 0.
  [Use the circle equation (x - h)2 + (y - k)2 = r2.]
Correct Answer is :   (x + 3)2 + (y - 5)2 = 0
Q12A circle is drawn with a diameter equal to the diagonal of a square whose side is 14 units. The center of the circle is (4, 7). Write the equation of the circle.
A. x2 + y2 - 8x - 14y - 98 = 0
B. x2 + y2 - 8x + 14y + 98 = 0
C. x2 + y2 - 8x - 14y - 33 = 0
D. x2 + y2 - 8x - 14y + 33 = 0

Solutions
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Step: 1
Side of the square = 14 units
Step: 2
Length of the diagonal = 2a, where a is the side of the square.
Step: 3
Diagonal of the square, d = 2 × 14 = 14 2units
Step: 4
Radius of the circle, r = d2 =1422 = 72 units.
  [Diameter of the circle is same as the diagonal of a square.]
Step: 5
Center : (h, k) = (4, 7)
Step: 6
Equation of the circle is (x - 4)2 + (y - 7)2 = (72)2
  [Use the formula (x - h)2 + (y - k)2 = r2.]
Step: 7
x2 - 8x + 16 + y2 - 14y + 49 = 98
Step: 8
x2 + y2 - 8x - 14y + 65 = 98
Step: 9
x2 + y2 - 8x - 14y - 33 = 0
Correct Answer is :   x2 + y2 - 8x - 14y - 33 = 0
Q13Find the equation of the circle passing through the points O(0, 0), A(14, 0) and B(0, 14).

A. x2 + y2 + 14x - 14y= 0
B. x2 + y2 + 14x + 14y = 0
C. x2 + y2 - 14x - 14y = 0
D. x2 + y2 - 14x + 14y = 0

Solutions
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Step: 1
The three points O(0, 0), A(14, 0) and B(0, 14) forms a right triangle.
Step: 2
In the right ΔOAB, the length of the hypotenuse = AB
Step: 3
= (0 - 14)2+(14 - 0)2 = 142
  [Distance formula.]
Step: 4
Midpoint of the hypotenuse, C = (14 + 02, 0 + 142) = (7, 7)
  [Use the mid point formula.]
Step: 5
The circle passing through O, A, B is the circum circle of Δ OAB whose center is at (7, 7) and radius is AB2 = 1422 = 72
Step: 6
So, the equation of the circle is (x - 7)2 + (y - 7)2 = (72)2.
  [Use (x - h)2 + (y - k)2 = r2.]
Step: 7
x2 + y2 - 14x - 14y = 0
  [Simplify.]
Correct Answer is :   x2 + y2 - 14x - 14y = 0
Q14If P (- 5, - 3) and Q (3, 5) are the end points of the diameter of a circle, then what is the equation of the circle?
A. (x + 3)2 + (y + 5)2 = 8
B. (x - 1)2 + (y + 1)2 = 32
C. (x + 1)2 + (y - 1)2 = 32
D. x2 + y2 = 32
E. (x + 4)2 + (y + 1)2 = 128

Solutions
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Step: 1
Diameter,PQ = (3+5)2+(5+3)2
  [Use distance formula.]
Step: 2
= (8)2+(8)2
Step: 3
= 64+64 = 128 = 82
  [Simplify.]
Step: 4
Diameter = 82 ⇒ radius = 822 = 42 units
Step: 5
Center of the circle = (-5+32, -3+52) = (- 1, 1)
  [Use midpoint formula and simplify.]
Step: 6
Equation of a circle with center (- 1, 1) and radius 42 units is (x + 1)2 + (y - 1)2 = (42)2
  [Use (x - h)2 + (y - k)2 = r2.]
Step: 7
So, the equation of the circle is (x + 1)2 + (y - 1)2 = 32.
Correct Answer is :   (x + 1)2 + (y - 1)2 = 32

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