# FIND CENTER AND RADIUS OF CIRCLE FROM STANDARD FORM

Equation of circle in standard form :

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

Here (-g, -f) is radius. and radius = √g2 + f2 - c

Note :

Using completing the square method, we can convert the given equation from standard form to (x - h)2 + (y - k)2 = r2, we get center and radius.

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 + 2x⋅5 + y2 – 2y⋅3 = -18

To complete the formula

x2 + 2x⋅5 + 52 - 52 + y2 – 2y⋅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 + 2x⋅9 + y2 – 2y⋅4 = -48

To complete the formula

x2 + 2x⋅9 + 92 - 92+ y2 – 2y⋅4 + 4- 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

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 - 2x⋅2 + y2 + 2y⋅3 = 113

x2 - 2x⋅2 + 22 - 22 + y2 + 2y⋅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).

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