Question

a circle has the equation x² + y² + 4x - 2y - 4 = 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 (-2,1) (type an ordered - pair, using integers or decimals.) the radius of the circle is 3 (type an integer or a decimal.) (b) use the graphing tool to graph the circle. click to enlarge graph edit coordinates (-8,1)

Explanation:

Step1: Rewrite the circle equation in standard form

The general equation of a circle is $x^{2}+y^{2}+4x - 2y-4 = 0$. 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 $(x + 2)^{2}-4+(y - 1)^{2}-1-4 = 0$, which simplifies to $(x + 2)^{2}+(y - 1)^{2}=9$.

Step2: Identify the center and radius

The standard - form of a circle equation is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center and $r$ is the radius.
Comparing $(x + 2)^{2}+(y - 1)^{2}=9$ with $(x - h)^{2}+(y - k)^{2}=r^{2}$, we have $h=-2,k = 1,r = 3$.

Step3: Find the $x$ - intercepts

Set $y = 0$ in the equation $(x + 2)^{2}+(y - 1)^{2}=9$.
$(x + 2)^{2}+(0 - 1)^{2}=9$, so $(x + 2)^{2}+1 = 9$, then $(x + 2)^{2}=8$, and $x+2=\pm\sqrt{8}=\pm2\sqrt{2}$.
So $x=-2\pm2\sqrt{2}$.

Step4: Find the $y$ - intercepts

Set $x = 0$ in the equation $(x + 2)^{2}+(y - 1)^{2}=9$.
$(0 + 2)^{2}+(y - 1)^{2}=9$, so $4+(y - 1)^{2}=9$, then $(y - 1)^{2}=5$, and $y - 1=\pm\sqrt{5}$.
So $y=1\pm\sqrt{5}$.

Answer:

(a) Center: $(-2,1)$; Radius: $3$
(b) To graph the circle, plot the center $(-2,1)$ and then use the radius of $3$ to draw the circle.
(c) $x$ - intercepts: $x=-2 + 2\sqrt{2}\approx0.83$ and $x=-2-2\sqrt{2}\approx - 4.83$; $y$ - intercepts: $y=1+\sqrt{5}\approx3.24$ and $y=1-\sqrt{5}\approx - 1.24$