For what value of $a$ does the equation $x ^ { 2 } + y ^ { 2 } - a x + 6 y + 33 = 0$ represent a circle with center $( 7 , - 3 )$ ? When $a$ assumes this value, what is the radius of the circle?
A) $\quad a = 14$, radius $= 5$
B) $\quad a = 13 , \quad$ radius $= 4$
C) $\quad a = 13$, radius $= 6$
D) $\quad a = 15 , \quad$ radius $= 7$
E) $\quad a = 12 , \quad$ radius $= 6$

Question

Quizplus · Accepted Answer

The general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Comparing this with the given equation $x^2 + y^2 - ax + 6y + 33 = 0$, and rearranging to match the general form, we get $(x^2 - ax) + (y^2 + 6y) = -33$. To complete the squares, we add and subtract $(\frac{a}{2})^2$ for the $x$ terms and add and subtract $(\frac{6}{2})^2 = 9$ for the $y$ terms, leading to $(x - \frac{a}{2})^2 + (y + 3)^2 = -33 + (\frac{a}{2})^2 + 9$. For the center to be $(7, -3)$, we need $\frac{a}{2} = 7$ and thus $a = 14$. Substituting $a = 14$ into the equation gives the radius squared as $-33 + 49 + 9 = 25$, so the radius $r = 5$.

For What Value Of aaa Does the Equation x2+y2−ax+6y+33=0x ^ { 2 } + y ^ { 2 } - a x + 6 y + 33 = 0x2+y2−ax+6y+33=0

For What Value Of $a$ Does the Equation $x ^ { 2 } + y ^ { 2 } - a x + 6 y + 33 = 0$