Rewrite the equation in standard form. Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 6 x - 8 y - 56 = 0$$
A) $( x + 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 81
center \(( - 3,4 ) , r = 9
B) \(( x - 3 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 81
center \(( 3 , - 4 ) , r = 9
C) \(( x - 4 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 81
center \(( 4 , - 3 ) , \ { r } = 9$
D) $( x + 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 81
center \(( - 3,4 ) , r = 81$

Question

Rewrite the equation in standard form. Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 6 x - 8 y - 56 = 0$$
A) $( x + 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 81
 center \(( - 3,4 ) , r = 9
B) \(( x - 3 ) ^ { 2 } + ( y + 4 ) ^ { 2 } = 81
 center \(( 3 , - 4 ) , r = 9
C) \(( x - 4 ) ^ { 2 } + ( y + 3 ) ^ { 2 } = 81
center \(( 4 , - 3 ) , \ { r } = 9$ 
D) $( x + 3 ) ^ { 2 } + ( y - 4 ) ^ { 2 } = 81
 center \(( - 3,4 ) , r = 81$

Quizplus · Accepted Answer

To rewrite the given equation in standard form, complete the square for both $x$ and $y$ terms: $$x^2 + 6x + y^2 - 8y = 56$$ becomes $$(x^2 + 6x + 9) + (y^2 - 8y + 16) = 56 + 9 + 16$$ which simplifies to $$(x + 3)^2 + (y - 4)^2 = 81$$. The center is at $(-3, 4)$ and the radius $r = \sqrt{81} = 9$.

Rewrite the Equation in Standard Form x2+y2+6x−8y−56=0x ^ { 2 } + y ^ { 2 } + 6 x - 8 y - 56 = 0x2+y2+6x−8y−56=0

Rewrite the Equation in Standard Form $x ^ { 2 } + y ^ { 2 } + 6 x - 8 y - 56 = 0$