Question

a circle has the equation x² + y²+4x - 2y - 20 = 0. (a) find the center (h,k) and radius r of the circle. (b) graph the circle. (c) find the intercepts, if any, of the graph. (a) the center of the circle is □ (type an ordered - pair, using integers or decimals.)

Explanation:

Step1: Rewrite the circle 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 $x^{2}+y^{2}+4x - 2y-20 = 0$. Complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}+4x=(x + 2)^{2}-4$.
For the $y$ - terms: $y^{2}-2y=(y - 1)^{2}-1$.
So the equation becomes $(x + 2)^{2}-4+(y - 1)^{2}-1-20 = 0$.

Step2: Simplify the equation to standard form

$(x + 2)^{2}+(y - 1)^{2}=20 + 4+1$.
$(x + 2)^{2}+(y - 1)^{2}=25$.

Step3: Identify the center and radius

Comparing with $(x - h)^2+(y - k)^2=r^2$, we have $h=-2,k = 1,r = 5$.

Answer:

(a) The center $(h,k)$ is $(-2,1)$.
(b) To graph the circle, plot the center $(-2,1)$ and then draw a circle with radius $r = 5$.
(c) x - intercepts: Set $y = 0$ in the equation $(x + 2)^{2}+(0 - 1)^{2}=25$.
$(x + 2)^{2}+1=25$, $(x + 2)^{2}=24$, $x+2=\pm\sqrt{24}=\pm2\sqrt{6}$, $x=-2\pm2\sqrt{6}$.
y - intercepts: Set $x = 0$ in the equation $(0 + 2)^{2}+(y - 1)^{2}=25$.
$4+(y - 1)^{2}=25$, $(y - 1)^{2}=21$, $y-1=\pm\sqrt{21}$, $y=1\pm\sqrt{21}$.