Question

a circle has the equation x² + y²+2x - 8y - 19 = 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 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 $x^{2}+y^{2}+2x - 8y-19 = 0$. Complete the square for $x$ and $y$ terms. For the $x$ - terms: $x^{2}+2x=(x + 1)^{2}-1$. For the $y$ - terms: $y^{2}-8y=(y - 4)^{2}-16$. So the equation becomes $(x + 1)^{2}-1+(y - 4)^{2}-16-19 = 0$, which simplifies to $(x + 1)^{2}+(y - 4)^{2}=36$.

Step2: Identify the center and radius

Comparing $(x + 1)^{2}+(y - 4)^{2}=36$ with $(x - h)^2+(y - k)^2=r^2$, we have $h=-1,k = 4,r = 6$. The center of the circle is $(-1,4)$.

Step3: For graphing

The center of the circle is at the point $(-1,4)$ and the radius is $r = 6$. To graph, plot the center $(-1,4)$ and then use the radius to find points on the circle by moving 6 units in all directions (up, down, left, right) from the center.

Step4: Find the intercepts

x - intercepts: Set $y = 0$ in the equation $(x + 1)^{2}+(y - 4)^{2}=36$. Then $(x + 1)^{2}+(0 - 4)^{2}=36$, so $(x + 1)^{2}+16 = 36$, $(x + 1)^{2}=20$, $x+1=\pm\sqrt{20}=\pm2\sqrt{5}$, $x=-1\pm2\sqrt{5}$.
y - intercepts: Set $x = 0$ in the equation $(x + 1)^{2}+(y - 4)^{2}=36$. Then $(0 + 1)^{2}+(y - 4)^{2}=36$, $(y - 4)^{2}=35$, $y-4=\pm\sqrt{35}$, $y=4\pm\sqrt{35}$.

Answer:

(a) The center of the circle is $(-1,4)$
(b) To graph: Plot the center $(-1,4)$ and draw a circle with radius $r = 6$.
(c) x - intercepts: $x=-1 + 2\sqrt{5},x=-1 - 2\sqrt{5}$; y - intercepts: $y=4+\sqrt{35},y=4-\sqrt{35}$