Question

a circle has the equation ( x^{2}+y^{2}+6x - 2y - 6 = 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 ( square ).
(type an ordered pair, using integers or decimals.)
the radius of the circle is ( square ).
(type an integer or a decimal.)
(b) use the graphing tool to graph the circle.

(c) what are the intercepts? select the correct choice and,
if necessary, fill in the answer box to complete your choice.
a. the intercept(s) is/are ( square ).
(type an ordered pair. use a comma to separate
answers as needed. type exact answers for
each coordinate, using radicals as needed.

Explanation:

Step1: Complete the square for \(x\) and \(y\) terms

Given the equation \(x^{2}+y^{2}+6x - 2y-6 = 0\).
Group \(x\) - terms and \(y\) - terms: \((x^{2}+6x)+(y^{2}-2y)=6\).
For the \(x\) - terms: \(x^{2}+6x=(x + 3)^{2}-9\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2b = 6\Rightarrow b = 3\)).
For the \(y\) - terms: \(y^{2}-2y=(y - 1)^{2}-1\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = y\), \(2b=2\Rightarrow b = 1\)).
Substitute back into the equation: \((x + 3)^{2}-9+(y - 1)^{2}-1=6\).
Simplify to get the standard form \((x + 3)^{2}+(y - 1)^{2}=16\).

Step2: Identify the center and radius

The standard form of a circle's equation is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Comparing \((x + 3)^{2}+(y - 1)^{2}=16=(x-(-3))^{2}+(y - 1)^{2}=4^{2}\).
So, \(h=-3\), \(k = 1\), and \(r = 4\).

Step3: Find the \(x\) - intercepts

Set \(y = 0\) in the equation \((x + 3)^{2}+(0 - 1)^{2}=16\).
\((x + 3)^{2}+1=16\).
\((x + 3)^{2}=15\).
\(x+3=\pm\sqrt{15}\).
\(x=-3\pm\sqrt{15}\). The \(x\) - intercepts are \((-3+\sqrt{15},0)\) and \((-3-\sqrt{15},0)\).

Step4: Find the \(y\) - intercepts

Set \(x = 0\) in the equation \((0 + 3)^{2}+(y - 1)^{2}=16\).
\(9+(y - 1)^{2}=16\).
\((y - 1)^{2}=7\).
\(y-1=\pm\sqrt{7}\).
\(y=1\pm\sqrt{7}\). The \(y\) - intercepts are \((0,1+\sqrt{7})\) and \((0,1-\sqrt{7})\).

Answer:

(a) The center of the circle is \((-3,1)\). The radius of the circle is \(4\).
(c) The intercept(s) is/are \((-3+\sqrt{15},0),(-3-\sqrt{15},0),(0,1+\sqrt{7}),(0,1 - \sqrt{7})\)