Question

1. what is the radius of the circle with the equation $(x - 10)^2+(y + 3)^2=1$?

a. 1
b. 0.5
c. 10
d. 2
what is the center of the circle with the equation $(x + 8)^2+(y - 6)^2=36$?
a. $(-8,6)$
b. $(8,-6)$
c. $(-8,-6)$
d. $(8,6)$

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 radius for the first equation \((x - 10)^2+(y + 3)^2 = 1\)

Comparing with \((x - h)^2+(y - k)^2=r^2\), we have \(r^2 = 1\). Taking the square root of both sides, \(r=\sqrt{1}=1\).

Step3: Find the center for the second equation \((x + 8)^2+(y - 6)^2=36\)

Rewrite \((x + 8)^2+(y - 6)^2=36\) as \((x-(-8))^2+(y - 6)^2=6^2\). Comparing with \((x - h)^2+(y - k)^2=r^2\), the center \((h,k)=(-8,6)\).

Answer:

For the radius question: a. \(1\)
For the center question: a. \((-8,6)\)