FIND CENTER AND RADIUS OF CIRCLE FROM STANDARD FORM

Problem 1 :

x² + 10x + y² – 6y = -18

The graph of the equation shown above is a circle. What is the radius of the circle?

A) 3    B) 4     C) 5    D) 9

Solution :

x² + 10x + y² – 6y = -18

x² + 2⋅x⋅5 + y² – 2⋅y⋅3 = -18

To complete the formula

x² + 2⋅x⋅5 + 5² - 5² + y² – 2⋅y⋅3 + 3² - 3² = -18

(x + 5)² + (y - 3)² - 25 - 9 = -18

(x + 5)² + (y - 3)² = -18 + 25 + 9

(x + 5)² + (y - 3)² = 16

(x + 5)² + (y - 3)² = 4²

(x - (-5))² + (y - 3)² = 4²

Comparing with

(x−h)² + (y−k)² = r²

(h, k) is (-5, 3) and radius = 4

So, option B is correct.

Problem 2 :

x² + 18x + y² – 8y = -48

The graph of the equation shown above is a circle. What is the radius of the circle?

A) 4    B) 5    C) 6     D) 7

Solution :

x² + 18x + y² – 8y = -48

x² + 2⋅x⋅9 + y² – 2⋅y⋅4 = -48

To complete the formula

x² + 2⋅x⋅9 + 9² - 9²+ y² – 2⋅y⋅4 + 4² - 4² = -48

(x + 9)² + (y - 4)² - 81 - 16 = -48

(x + 9)² + (y - 4)² - 97 = -48

(x - (-9))² + (y - 4)² = -48 + 97

(x - (-9))² + (y - 4)² = 49

(x - (-9))² + (y - 4)² = 7²

Radius = 7

Problem 3 :

x² - 4x + y² + 6y = 113

The graph of the equation shown above is a circle. What is the coordinate point of the center of the circle?

A) (13, 10)    B) (4, 13)     C) (-4, 6)     D) (2, -3)

Solution :

x² - 4x + y² + 6y = 113

x² - 2⋅x⋅2 + y² + 2⋅y⋅3 = 113

x² - 2⋅x⋅2 + 2² - 2² + y² + 2⋅y⋅3 + 3² - 3² = 113

(x - 2)² - 4 + (y + 3)² - 9 = 113

(x - 2)² + (y + 3)² - 13 = 113

(x - 2)² + (y + 3)² = 113 + 13

(x - 2)² + (y + 3)² = 126

h = 2 and k = -3

So, the center of the circle is (2, -3).