Question

what is the radius of the circle with the equation $(x + 5)^{2}+(y - 6)^{2}=49$?
a. 5
b. 49
c. 14
d. 7
which of the following is the equation of a circle with center $(-5,-5)$ and radius 10?
a. $(x - 5)^{2}+(y - 5)^{2}=100$
b. $(x + 5)^{2}+(y + 5)^{2}=20$
c. $(x + 5)^{2}+(y + 5)^{2}=100$
d. $(x - 5)^{2}+(y - 5)^{2}=10$
if the equation of a circle is $x^{2}+y^{2}=144$, what is the radius of the circle?
a. 36
b. 12
c. 72
d. 144
what is the radius of the circle with the equation $(x - 10)^{2}+(y + 3)^{2}=1$?
a. 0.5
b. 2
c. 1
d. 4

Explanation:

Step1: Recall circle standard form

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.

Step2: Find radius for first question

Given equation: $(x+5)^2 + (y-6)^2 = 49$. Here, $r^2 = 49$, so $r = \sqrt{49} = 7$.

Step3: Find equation for second question

Center $(-5,-5)$ means $h=-5$, $k=-5$; radius 10 means $r^2=100$. Substitute into standard form: $(x - (-5))^2 + (y - (-5))^2 = 100$, which simplifies to $(x+5)^2 + (y+5)^2 = 100$.

Step4: Find radius for third question

Given equation: $x^2 + y^2 = 144$. This is $(x-0)^2 + (y-0)^2 = 144$, so $r^2=144$, $r=\sqrt{144}=12$.

Step5: Find radius for fourth question

Given equation: $(x-10)^2 + (y+3)^2 = 1$. Here, $r^2=1$, so $r=\sqrt{1}=1$.

Answer:

1. d. 7
2. c. $(x + 5)^2 + (y + 5)^2 = 100$
3. b. 12
4. c. 1