Identify the center and radius of the circle. $$x ^ { 2 } + 9 x + y ^ { 2 } - \frac { 3 } { 2 } y = 10$$
A) Center: $\left( \frac { 9 } { 2 } , \frac { 3 } { 2 } \right) ; r = 10$
B) Center: $\left( - \frac { 9 } { 2 } , \frac { 3 } { 4 } \right) ; r = \frac { \sqrt { 493 } } { 4 }$
C) Center: $\left( - 9 , \frac { 3 } { 2 } \right) ; r = \sqrt { 10 }$
D) Center: $\left( \frac { 3 } { 4 } , - \frac { 1 } { 16 } \right) ; r = 2 \sqrt { 2 }$

Question

Quizplus · Accepted Answer

To find the center and radius of the circle given by the equation $x^2 + 9x + y^2 - \frac{3}{2}y = 10$, we need to complete the square for both $x$ and $y$ terms. For $x^2 + 9x$, we add and subtract $\left(\frac{9}{2}\right)^2 = \frac{81}{4}$. For $y^2 - \frac{3}{2}y$, we add and subtract $\left(\frac{3}{4}\right)^2 = \frac{9}{16}$. Adding these to both sides, we get: $$x^2 + 9x + \frac{81}{4} + y^2 - \frac{3}{2}y + \frac{9}{16} = 10 + \frac{81}{4} + \frac{9}{16}$$Simplifying, we get:$$\left(x + \frac{9}{2}\right)^2 + \left(y - \frac{3}{4}\right)^2 = \frac{493}{16}$$Thus, the center is $\left(-\frac{9}{2}, \frac{3}{4}\right)$ and the radius $r = \sqrt{\frac{493}{16}} = \frac{\sqrt{493}}{4}$.

Identify the Center and Radius of the Circle x2+9x+y2−32y=10x ^ { 2 } + 9 x + y ^ { 2 } - \frac { 3 } { 2 } y = 10x2+9x+y2−23​y=10

Identify the Center and Radius of the Circle $x ^ { 2 } + 9 x + y ^ { 2 } - \frac { 3 } { 2 } y = 10$