Question

(a)
the equation of a circle c is $(x + 2)^2+(y - 7)^2 = 36$. what is its center $(h,k)$?
a $(-2,-7)$
b $(-2,7)$
c $(2,-7)$
d $(2,7)$
(b)
part b
what is the radius $(r)$ of circle c?
answer: $r=$

Explanation:

Part (a)

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 of the circle.

Step2: Rewrite the given equation to match the standard form

The given equation is \((x + 2)^2+(y - 7)^2 = 36\), which can be rewritten as \((x-(- 2))^2+(y - 7)^2=6^2\).
Comparing with \((x - h)^2+(y - k)^2=r^2\), we get \(h=-2\) and \(k = 7\).

Part (b)

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 \(r\) is the radius of the circle.

Step2: Identify the radius from the given equation

The given equation is \((x + 2)^2+(y - 7)^2 = 36\), which can be rewritten as \((x-(- 2))^2+(y - 7)^2=6^2\).
Comparing with \((x - h)^2+(y - k)^2=r^2\), we get \(r^2 = 36\). Taking the square - root of both sides (since \(r>0\) for the radius of a circle), \(r=\sqrt{36}=6\).

Answer:

(a) B. \((-2,7)\)
(b) \(r = 6\)