Question

the equation of a circle is ( x^2 + 8x + y^2 - 12y = 144 ). what are the coordinates of the center and the length of the radius of the circle? a center (4, -6) and radius 12; b center (-4, 6) and radius 12; c center (4, -6) and radius 14; d center (-4, 6) and radius 14.

Explanation:

Step1: Complete the square for x-terms

The given equation is \(x^{2}+8x + y^{2}-12y=144\). For the \(x\)-terms, take the coefficient of \(x\) which is 8, divide by 2 to get 4, and square it to get \(4^{2} = 16\). Add 16 to both sides.

Step2: Complete the square for y-terms

For the \(y\)-terms, the coefficient of \(y\) is - 12, divide by 2 to get - 6, square it to get \((-6)^{2}=36\). Add 36 to both sides.
Now the equation becomes:
\(x^{2}+8x + 16+y^{2}-12y + 36=144 + 16+36\)

Step3: Rewrite as standard circle equation

The left - hand side can be written as \((x + 4)^{2}+(y-6)^{2}\) (using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\) and \((a - b)^{2}=a^{2}-2ab + b^{2}\)), and the right - hand side is \(144+16 + 36=196\).
The standard form of the equation of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Comparing \((x + 4)^{2}+(y - 6)^{2}=196\) with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-4\), \(k = 6\) and \(r^{2}=196\), so \(r=\sqrt{196}=14\).

Answer:

D. center \((-4,6)\) and radius 14