Question

rewrite the following equation in the center - radius form of the equation of a circle.
$x^{2}-16x + y^{2}+16y=-112$
a. $(x - 10)^{2}+(y + 7)^{2}=16$
b. $(x - 8)^{2}+(y + 8)^{2}=16$
c. $(x - 8)^{2}+(y + 9)^{2}=16$
d. $(x + 8)^{2}+(y - 8)^{2}=16$

Explanation:

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

Given \(x^{2}-16x + y^{2}+16y=-112\).
For \(x\) terms: \(x^{2}-16x=(x - 8)^{2}-64\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = x\), \(2b = 16\Rightarrow b = 8\)).

Step2: Complete the square for \(y\) terms

For \(y\) terms: \(y^{2}+16y=(y + 8)^{2}-64\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a = y\), \(2b = 16\Rightarrow b = 8\)).

Step3: Substitute back into the equation

Substitute \(x^{2}-16x=(x - 8)^{2}-64\) and \(y^{2}+16y=(y + 8)^{2}-64\) into \(x^{2}-16x + y^{2}+16y=-112\).
We get \((x - 8)^{2}-64+(y + 8)^{2}-64=-112\).

Step4: Simplify the equation

\((x - 8)^{2}+(y + 8)^{2}-128=-112\).
Add \(128\) to both sides: \((x - 8)^{2}+(y + 8)^{2}=-112 + 128\).
So \((x - 8)^{2}+(y + 8)^{2}=16\).

Answer:

B. \((x - 8)^{2}+(y + 8)^{2}=16\)