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 + 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).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM