Question

for the equation $x^{2}+y^{2}-8x - 4y-16 = 0$, do the following. (a) find the center $(h,k)$ and radius $r$ of the circle. (b) graph the circle. (c) find the intercepts, if any. (a) the center is (type an ordered pair.)

Explanation:

Step1: Rewrite the equation in standard form

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. Given $x^{2}+y^{2}-8x - 4y-16 = 0$. Complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}-8x=(x - 4)^{2}-16$.
For the $y$ - terms: $y^{2}-4y=(y - 2)^{2}-4$.
So the equation becomes $(x - 4)^{2}-16+(y - 2)^{2}-4-16 = 0$.

Step2: Simplify the equation

$(x - 4)^{2}+(y - 2)^{2}=16 + 4+16$.
$(x - 4)^{2}+(y - 2)^{2}=36$.

Step3: Identify the center and radius

Comparing with $(x - h)^2+(y - k)^2=r^2$, we have $h = 4$, $k = 2$ and $r = 6$.

Answer:

(a) $(4,2)$
(b) To graph the circle, plot the center $(4,2)$ and then use the radius $r = 6$ to draw the circle around the center.
(c)
x - intercepts: Set $y = 0$ in the equation $(x - 4)^{2}+(0 - 2)^{2}=36$.
$(x - 4)^{2}+4 = 36$.
$(x - 4)^{2}=32$.
$x-4=\pm\sqrt{32}=\pm4\sqrt{2}$.
$x=4\pm4\sqrt{2}$. So the x - intercepts are $(4 + 4\sqrt{2},0)$ and $(4-4\sqrt{2},0)$.
y - intercepts: Set $x = 0$ in the equation $(0 - 4)^{2}+(y - 2)^{2}=36$.
$16+(y - 2)^{2}=36$.
$(y - 2)^{2}=20$.
$y-2=\pm\sqrt{20}=\pm2\sqrt{5}$.
$y=2\pm2\sqrt{5}$. So the y - intercepts are $(0,2 + 2\sqrt{5})$ and $(0,2-2\sqrt{5})$.