Question

this equation describes a circle.
find the center and radius of the circle. show your work.
$x^{2}+4x + y^{2}-8y - 5 = 0$
a. center $(4,-8)$ radius $=5$
b. center $(-2,4)$ radius $=10$
c. center $(-2,4)$ radius $=5$
d. center $(4,-8)$ radius $=10$
e. center $(-2,4)$ radius $=25$
f. center $(4,-8)$ radius $=25$
question: 6
a circle drawn on a coordinate plane has the $x^{2}+y^{2}+8x - 8y + 24 = 0$
a. the center is located at point $(-4,4)$
b. the center is located at point $(4,-4)$
c. the center is located at point $(-8,8)$
d. the center is located at point $(8,-8)$

Explanation:

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

For the equation \(x^{2}+4x + y^{2}-8y-5 = 0\), for the \(x\) - terms \(x^{2}+4x=(x + 2)^{2}-4\) (using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=x\), \(2b = 4\Rightarrow b = 2\)).

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

For the \(y\) - terms \(y^{2}-8y=(y - 4)^{2}-16\) (using the formula \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a = y\), \(2b=8\Rightarrow b = 4\)).

Step3: Rewrite the equation of the circle

Substitute the completed - square expressions into the original equation:

$$
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$$

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
Comparing \((x + 2)^{2}+(y - 4)^{2}=25=(x-(-2))^{2}+(y - 4)^{2}=5^{2}\), the center \((h,k)=(-2,4)\) and radius \(r = 5\).

Answer:

C. Center \((-2,4)\) Radius \(=5\)