Question

question: 5
this equation describes a circle.
find the center and radius of the circle. show your work.
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 equation ( 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:

Question 5

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

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

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

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

Step3: Rewrite the equation

Substitute into the original equation: \((x + 2)^{2}-4+(y - 4)^{2}-16-5=0\).
Simplify to \((x + 2)^{2}+(y - 4)^{2}=25\).
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.
Here \(h=-2\), \(k = 4\), \(r^{2}=25\Rightarrow r = 5\).

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

Given equation \(x^{2}+y^{2}+8x-8y + 24=0\).
For \(x\) terms: \(x^{2}+8x=(x + 4)^{2}-16\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), \(a=x\), \(2b = 8\Rightarrow b=4\)).

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

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

Step3: Rewrite the equation

Substitute into the original equation: \((x + 4)^{2}-16+(y - 4)^{2}-16+24=0\).
Simplify: \((x + 4)^{2}+(y - 4)^{2}=8\).
The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center. Here \(h=-4\), \(k = 4\).

Answer:

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

Question 6