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.

Problem 9 :

Two circles have equations

(x + 1)² + (y + 3)² = 20 and x² + y² – 10x – 18y + 26 = 0

(a) Write down the centre and radius of each circle.

(b) Show that the circles touch at a single point.

(c) Find P, the point of contact of the circles.

Solution :

a)

Finding center and radius of the circle (x + 1)² + (y + 3)² = 20 :

(x - (-1))² + (y - (-3))² = 20

Center (h, k) ==> (-1, -3)

Radius = √20

Finding the center and radius of x² + y² – 10x – 18y + 26 = 0 :

x² – 10x + y² – 18y + 26 = 0

x² – 2x(5) + 5² - 5²+ y² – 2y(9) + 9² - 9² + 26 = 0

(x - 5)² - 25 + (y - 9)² – 81 + 26 = 0

(x - 5)² + (y - 9)² – 106 + 26 = 0

(x - 5)² + (y - 9)² - 80 = 0

(x - 5)² + (y - 9)² = 80

Center (h, k) ==> (5, 9)

Radius = √80

(b) To prove the circles are touching each other at single point, we should prove

the distance between two centers = sum of the radii of two circles

= √(x₂ - x₁)² + (y₂ - y₁)²

(-1, -3) and (5, 9)

= √(5 + 1)² + (9 + 3)²

= √6² + 12²

= √(36 + 144)

=  √180

= √(2 x 2 x 5 x 3 x 3)

= 6√5 ----(1)

Sum of the radii = √20 + √80

= 2√5 + 4√5

= 6√5 ----(2)

(c) Find P, the point of contact of the circles.

Let (x, y) be the required point

The point (x, y) is dividing the line segment in the ratio of 2√5 : 4√5

= 2 : 4

= 1 : 2

= (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)

(-1, -3) and (5, 9)

[1(5) + 2(-1)]/(1 + 2), [1(9) + 2(-3)]/(1 + 2) = (x, y)

3/3, 3/3 = (x, y)

(x, y) = (1, 1)

So, the required point is (1, 1).