# HOW TO FIND CENTER AND RADIUS OF A CIRCLE FROM AN EQUATION

Equation of circle

x2 + y2 = r2

Center (0, 0) and radius is r.

(x - h)2 + (y - k)2 = r2

Center (h, k) and radius is r.

x2 + y2 + 2gx + 2fy + c = 0

Center (-g, -f) and radius is r = √(g2 + f2 – c)

Write down the center and radius of each circle below

Problem 1 :

x2 + y2 = 25

Solution :

The given equation of the circle is in the form of

x2 + y2 = r2

So, the center of the given circle is (0, 0).

r2 = 25

r = 5 units

Problem 2 :

x2 + y2 = 12

Solution :

The given equation of the circle is in the form of

x2 + y2 = r2

So, the center of the given circle is (0, 0).

r2 = 12

r = √12 units

Problem 3 :

(x – 3)2 + (y – 2)2 = 36

Solution :

The given equation of the circle is in the form of

(x – h)2 + (y – k)2 = r2

Center :

(h, k) = (3, 2)

r2 = 36

r = 6 units

Problem 4 :

(x + 1)2 + (y – 4)2 = 10

Solution :

The given equation of the circle is in the form of

(x – h)2 + (y – k)2 = r2

Center :

(h, k) = (-1, 4)

r2 = 10

r = √10 units

Problem 5 :

x2 + y2 – 10x – 6y - 2 = 0

Solution :

The given equation of circle is in general form.

Comparing

x2 + y2 – 10x – 6y - 2 = 0

Using the completing the square method,

x2 – 2 · x · 5 + 52 - 52 + y2 - 2 ·y·3 + 32 - 32 - 2 = 0

(x - 5)2 + (y - 3)2 - 25 - 9 - 2 = 0

(x - 5)2 + (y - 3)2 - 36 = 0

(x - 5)2 + (y - 3)2 = 36

Comparing with (x - h)2 + (y - k)2 = r2

Center :

(-g, -f) = (5, 3)

Problem 6 :

x2 + y2 + 6x + 4y + 4 = 0

Solution :

The given equation of circle is in general form.

Comparing

x2 + y2 + 6x + 4y + 4 = 0

Using the completing the square method,

x2 + 2 · x · 3 + 32 - 32 + y2 + 2 ·y·2 + 22 - 22 + 4 = 0

(x + 3)2 + (y + 2)2 - 9  - 4 + 4 = 0

(x + 3)2 + (y + 2)2 - 9 = 0

(x + 3)2 + (y + 2)2 = 9

Center :

(-g, -f) = (-3, -2)

Problem 7 :

The point (a, 5) lies on the circle with equation x2 + y2 = 74. Find two values for a.

Solution :

Given, x2 + y2 = 74

The point (a, 5) = (x, y)

(a)2 + (5)2 = 74

(a)2 + 25 = 74

a2 = 74 – 25

a2 = 49

a = ±7

So, two values for a is 7, -7.

Problem 8 :

The point (3, c) lies on the circle x2 + y2 – 4x + 6y + 12 = 0. Find c.

Solution :

Given, x2 + y2 – 4x + 6y + 12 = 0

The point (3, c) = (x, y)

(3)2 + (c)2 – 4(3) + 6(c) + 12 = 0

9 + c2 – 12 + 6c + 12 = 0

9 + c2 + 6c = 0

c2 + 6c + 9 = 0

(c + 3) (c + 3) = 0

(c + 3)2 = 0

c + 3 = 0

c = -3

So, c is -3.

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