QUESTION IMAGE
Question
a circle has the equation ( x^{2}+y^{2}+2x - 6y - 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.
oa. 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. simplify your answer.)
Step1: Complete the square for \(x\) and \(y\) terms
Given equation \(x^{2}+y^{2}+2x - 6y-6 = 0\)
Group \(x\) terms and \(y\) terms: \((x^{2}+2x)+(y^{2}-6y)=6\)
For \(x\): \((x^{2}+2x+1)-1\) (since \((a + b)^2=a^{2}+2ab + b^{2}\), here \(a=x\), \(b = 1\))
For \(y\): \((y^{2}-6y + 9)-9\) (since \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=y\), \(b = 3\))
The equation becomes \((x + 1)^{2}-1+(y - 3)^{2}-9=6\)
Step2: Write in standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\)
Simplify the equation: \((x + 1)^{2}+(y - 3)^{2}=6 + 1+9\)
\((x + 1)^{2}+(y - 3)^{2}=16\)
So, \(h=-1,k = 3,r = 4\)
Step3: Find \(x\) - intercepts (\(y = 0\))
Substitute \(y = 0\) into \((x + 1)^{2}+(y - 3)^{2}=16\)
\((x + 1)^{2}+(0 - 3)^{2}=16\)
\((x + 1)^{2}+9=16\)
\((x + 1)^{2}=7\)
\(x+1=\pm\sqrt{7}\)
\(x=-1\pm\sqrt{7}\)
\(x\) - intercepts: \((-1+\sqrt{7},0),(-1-\sqrt{7},0)\)
Step4: Find \(y\) - intercepts (\(x = 0\))
Substitute \(x = 0\) into \((x + 1)^{2}+(y - 3)^{2}=16\)
\((0 + 1)^{2}+(y - 3)^{2}=16\)
\(1+(y - 3)^{2}=16\)
\((y - 3)^{2}=15\)
\(y-3=\pm\sqrt{15}\)
\(y=3\pm\sqrt{15}\)
\(y\) - intercepts: \((0,3+\sqrt{15}),(0,3-\sqrt{15})\)
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(a) The center of the circle is \((-1,3)\). The radius of the circle is \(4\).
(c) The intercepts are \((-1+\sqrt{7},0),(-1-\sqrt{7},0),(0,3+\sqrt{15}),(0,3-\sqrt{15})\)