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

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).

Radius :

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).

Radius :

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)

Radius :

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)

Radius :

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)

Radius :

Radius = 6 units

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)

Radius :

Radius = 3 units

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.