Find the center (h, k) and radius r of the circle with the given equation.
-$$4 x ^ { 2 } + 4 y ^ { 2 } - 12 x + 16 y - 5 = 0$$
A) $( \mathrm { h } , \mathrm { k } ) = \left( \frac { 3 } { 2 } , - 2 \right) ; \mathrm { r } = \frac { \sqrt { 30 } } { 2 }$
B) $( \mathrm { h } , \mathrm { k } ) = \left( - \frac { 3 } { 2 } , 2 \right) ; \mathrm { r } = \frac { 3 \sqrt { 5 } } { 2 }$
C) $( \mathrm { h } , \mathrm { k } ) = \left( - \frac { 3 } { 2 } , 2 \right) ; \mathrm { r } = \frac { \sqrt { 30 } } { 2 }$
D) $( \mathrm { h } , \mathrm { k } ) = \left( \frac { 3 } { 2 } , - 2 \right) ; \mathrm { r } = \frac { 3 \sqrt { 5 } } { 2 }$

Question

Quizplus · Accepted Answer

To find the center and radius, we complete the square for both $x$ and $y$ terms in the given equation $4x^2 + 4y^2 - 12x + 16y - 5 = 0$. First, factor out the 4 from the $x$ and $y$ terms: $4(x^2 - 3x) + 4(y^2 + 4y) - 5 = 0$. Completing the square for $x^2 - 3x$ involves adding and subtracting $(\frac{3}{2})^2 = \frac{9}{4}$, and for $y^2 + 4y$ adding and subtracting $(2)^2 = 4$. After adjusting for the factor of 4 outside, we add $4 \times \frac{9}{4} = 9$ and $4 \times 4 = 16$ inside the equation and compensate by adding them to the other side as well. This gives us $4(x^2 - 3x + \frac{9}{4}) + 4(y^2 + 4y + 4) = 5 + 9 + 16$, simplifying to $4(x - \frac{3}{2})^2 + 4(y + 2)^2 = 30$. Dividing by 4, we get $(x - \frac{3}{2})^2 + (y + 2)^2 = \frac{30}{4}$, which shows the center $(h, k) = (\frac{3}{2}, -2)$ and radius $r = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2}$.

Find the Center (H, K) and Radius R of the Circle