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 (type an integer or a decimal.)

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 \(x\) - terms: \(x^{2}+4x=(x + 2)^{2}-4\).
For \(y\) - 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 \(h=-2\), \(k = 1\), and \(r^{2}=9\), so \(r = 3\).

Step3: Find the \(x\) - intercepts

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

Step4: Find the \(y\) - intercepts

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

Answer:

(a) Center: \((-2,1)\)
Radius: \(3\)
(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 \(y=1 - \sqrt{5}\approx - 1.24\)