Question

which radius corresponds to the circle with the equation (x + 5)^2+(y - 1)^2 = 121?
a. 13
b. 12
c. 11
d. 121
what is the equation of a circle with center (-2,1) and radius 6?
a. (x - 2)^2+(y + 1)^2 = 6
b. (x + 2)^2+(y + 1)^2 = 36
c. (x + 2)^2+(y - 1)^2 = 36
d. (x - 2)^2+(y + 1)^2 = 36
what is the standard form of the circle with center (4,-3) and radius 5?
a. (x + 4)^2+(y - 3)^2 = 25
b. (x - 4)^2+(y + 3)^2 = 25
c. (x + 4)^2+(y - 3)^2 = 5
d. (x - 4)^2+(y + 3)^2 = 5

Explanation:

Step1: Recall circle - equation formula

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

First question

The equation of the circle is $(x + 5)^2+(y - 1)^2=121$. Comparing with the standard - form $(x - h)^2+(y - k)^2=r^2$, we have $r^2 = 121$.

Step2: Solve for the radius

Take the square - root of both sides of the equation $r^2 = 121$. Since $r>0$, $r=\sqrt{121}=11$.

Second question

The center of the circle is $(h,k)=(-2,1)$ and the radius $r = 6$. Substitute these values into the standard - form $(x - h)^2+(y - k)^2=r^2$.
We get $(x-(-2))^2+(y - 1)^2=6^2$, which simplifies to $(x + 2)^2+(y - 1)^2=36$.

Third question

The center of the circle is $(h,k)=(4,-3)$ and the radius $r = 5$. Substitute into the standard - form $(x - h)^2+(y - k)^2=r^2$.
We have $(x - 4)^2+(y-(-3))^2=5^2$, which simplifies to $(x - 4)^2+(y + 3)^2=25$.

Answer:

1. C. 11
2. C. $(x + 2)^2+(y - 1)^2=36$
3. B. $(x - 4)^2+(y + 3)^2=25$