Write the equation in standard form to find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 6 x + 14 y + 37 = 0$$
A) $( x + 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 } = 21 ;$ center $( - 3 , - 7 )$, radius $\sqrt { 21 }$
B) $( x + 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 } = 21 ;$ center $( - 3 , - 7 )$, radius 21
C) $( x + 3 ) ^ { 2 } + ( y + 7 ) ^ { 2 } = 21 ;$ center $( 3,7 )$, radius 21
D) $( x - 3 ) ^ { 2 } + ( y - 7 ) ^ { 2 } = 21 ;$ center $( 3,7 )$, radius $\sqrt { 21 }$

Question

Quizplus · Accepted Answer

To convert the given equation to the standard form of a circle $(x - h)^2 + (y - k)^2 = r^2$, we complete the square for both $x$ and $y$ terms. For $x^2 + 6x$, we add and subtract $(\frac{6}{2})^2 = 9$, and for $y^2 + 14y$, we add and subtract $(\frac{14}{2})^2 = 49$. The equation becomes $x^2 + 6x + 9 + y^2 + 14y + 49 = 0 + 9 + 49$, simplifying to $(x + 3)^2 + (y + 7)^2 = 58 - 37 = 21$. The center is at $(-3, -7)$ and the radius is $\sqrt{21}$.

Write the Equation in Standard Form to Find the Center x2+y2+6x+14y+37=0x ^ { 2 } + y ^ { 2 } + 6 x + 14 y + 37 = 0x2+y2+6x+14y+37=0

Write the Equation in Standard Form to Find the Center $x ^ { 2 } + y ^ { 2 } + 6 x + 14 y + 37 = 0$