Question

determine each of the following for the circle whose equation is $x^{2}+y^{2}+10y - 11 = 0$. make sure your answer is fully simplified.
the equation of the circle in standard (center, radius) form is:
the center of the circle is:
the radius of the circle is:
graph the circle:

Explanation:

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

Given \(x^{2}+y^{2}+10y - 11=0\).
Group the \(y\) terms: \(x^{2}+(y^{2}+10y)=11\).
For \(y^{2}+10y\), using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a = y\) and \(2b = 10\) (so \(b = 5\)). Then \(y^{2}+10y=(y + 5)^{2}-25\).
The equation becomes \(x^{2}+(y + 5)^{2}-25=11\).

Step2: Write in standard form

Add \(25\) to both sides of the equation \(x^{2}+(y + 5)^{2}-25=11\).
We get \(x^{2}+(y + 5)^{2}=11 + 25\).
So \(x^{2}+(y+5)^{2}=36\).

Answer:

• The equation of the circle in standard (center, radius) form is: \(x^{2}+(y + 5)^{2}=36\)
• The center of the circle is: \((0,-5)\)
• The radius of the circle is: \(6\)