Question

what is the equation of a circle with center at (2, - 5) and radius 3? a. (x - 2)^2+(y + 5)^2 = 9 b. (x + 2)^2+(y - 5)^2 = 9 c. (x + 2)^2+(y - 5)^2 = 3 d. (x - 2)^2+(y + 5)^2 = 81 which of the following is the equation of a circle with center (0, - 1) and radius 6? a. x^2 + y^2 = 36 b. x^2+(y + 1)^2 = 12 c. x^2 + y^2 = 12 d. x^2+(y + 1)^2 = 36 what is the radius of the circle with the equation (x + 5)^2+(y - 6)^2 = 49? a. 7 b. 49 c. 14 d. 5

Explanation:

Step1: Recall circle - equation formula

The standard form of the equation of a circle 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 center of the circle is $(2,-5)$ and the radius $r = 3$. Substitute $h = 2$, $k=-5$ and $r = 3$ into the standard - form equation.
We get $(x - 2)^2+(y+5)^2=3^2=(x - 2)^2+(y + 5)^2=9$. So the answer to the first question is a.

Second question

The center of the circle is $(0,-1)$ and the radius $r = 6$. Substitute $h = 0$, $k=-1$ and $r = 6$ into the standard - form equation.
We have $(x - 0)^2+(y + 1)^2=6^2$, which simplifies to $x^2+(y + 1)^2=36$. So the answer to the second question is d.

Third question

The equation of the circle is $(x + 5)^2+(y - 6)^2=49$. Comparing it with the standard form $(x - h)^2+(y - k)^2=r^2$, we know that $r^2=49$.
Take the square root of both sides. Since $r>0$, $r=\sqrt{49}=7$. So the answer to the third question is a.

Answer:

1. a. $(x - 2)^2+(y + 5)^2=9$
2. d. $x^2+(y + 1)^2=36$
3. a. 7