Question

identify the center of the circle given by the equation $(x - 6)^2+(y + 7)^2 = 64$.
a. $(6,-7)$
b. $(-6,7)$
c. $(-6,-7)$
d. $(6,7)$
what is the radius of the circle with the equation $(x + 2)^2+(y - 3)^2 = 100$?
a. 20
b. 10
c. 5
d. 4

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of a circle's equation is \((x - h)^2+(y - k)^2=r^2\), where \((h,k)\) is the center and \(r\) is the radius.

Step2: Identify the center for the first equation \((x - 6)^2+(y + 7)^2 = 64\)

Compare with \((x - h)^2+(y - k)^2=r^2\). Here \(h = 6\), \(k=-7\) (since \(y+7=y-(-7)\)). So the center is \((6,-7)\).

Step3: Find the radius for the second equation \((x + 2)^2+(y - 3)^2=100\)

Compare with \((x - h)^2+(y - k)^2=r^2\). We have \(r^2 = 100\). Taking the square root of both sides (\(r>0\)), \(r=\sqrt{100}=10\).

Answer:

For the center of the circle \((x - 6)^2+(y + 7)^2 = 64\): a. \((6,-7)\)
For the radius of the circle \((x + 2)^2+(y - 3)^2=100\): b. \(10\)