Question

what is the center of the circle with the equation $(x - 7)^2+(y + 8)^2 = 49$?
a. $(-7,-8)$
b. $(7,8)$
c. $(-7,8)$
d. $(7,-8)$
what is the radius of the circle with the equation $(x - 3)^2+(y + 4)^2 = 16$?
a. 3
b. 4
c. 16
d. 8

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: Find the center of the first circle

For the equation \((x - 7)^2+(y + 8)^2 = 49\), rewrite \(y+8\) as \(y-(-8)\). Comparing with \((x - h)^2+(y - k)^2=r^2\), we have \(h = 7\) and \(k=-8\). So the center is \((7,-8)\).

Step3: Find the radius of the second circle

For the equation \((x - 3)^2+(y + 4)^2=16\), rewrite it as \((x - 3)^2+(y -(-4))^2 = 4^2\). Comparing with \((x - h)^2+(y - k)^2=r^2\), we get \(r = 4\).

Answer:

For the center question: d. \((7,-8)\)
For the radius question: b. \(4\)