Find the center and radius of the circle.
-$$x ^ { 2 } + y ^ { 2 } + 10 x + 14 y + 38 = 0$$
A) center: $( - 5 , - 7 )$; radius: 6
B) center: $( 5,7 )$; radius: 36
C) center: $( 7,5 )$; radius: 36
D) center: $( - 7 , - 5 )$; radius: 6

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + y^2 + 10x + 14y + 38 = 0$, we complete the square for both $x$ and $y$. The general form of a circle's equation is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.1. Group $x$ and $y$ terms: $(x^2 + 10x) + (y^2 + 14y) = -38$.2. Complete the square for $x$ and $y$:   - For $x^2 + 10x$, add and subtract $(\frac{10}{2})^2 = 25$.   - For $y^2 + 14y$, add and subtract $(\frac{14}{2})^2 = 49$.3. The equation becomes $x^2 + 10x + 25 + y^2 + 14y + 49 = -38 + 25 + 49$.4. Simplify: $(x + 5)^2 + (y + 7)^2 = 36$.5. The center is $(-5, -7)$ and the radius is $\sqrt{36} = 6$.

Find the Center and Radius of the Circle x2+y2+10x+14y+38=0x ^ { 2 } + y ^ { 2 } + 10 x + 14 y + 38 = 0x2+y2+10x+14y+38=0

Find the Center and Radius of the Circle $x ^ { 2 } + y ^ { 2 } + 10 x + 14 y + 38 = 0$