Question

what is the center of a circle with the equation x^2 + y^2 + 10x - 6y = 35?
a. (5,3)
b. (-5,-3)
c. (-5,3)
d. (5,-3)

if the center of a circle is (4, -3) and its radius is 8, what is the equation?
a. (x - 4)^2 + (y + 3)^2 = 4^2
b. (x - 4)^2 + (y + 3)^2 = 16
c. (x + 4)^2 + (y - 3)^2 = 64
d. (x - 4)^2 + (y + 3)^2 = 64

a circle has a center at (3, -4) and passes through (6, -4). what is its equation?
a. (x - 3)^2 + (y + 4)^2 = 9
b. (x - 3)^2 + (y + 4)^2 = 6
c. (x + 3)^2 + (y - 4)^2 = 9
d. (x + 3)^2 + (y + 4)^2 = 3^2

Explanation:

Step1: Recall circle - equation form

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. For the equation $x^{2}+y^{2}+10x - 6y = 35$, we complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}+10x=(x + 5)^{2}-25$. For the $y$ - terms: $y^{2}-6y=(y - 3)^{2}-9$.
So the equation becomes $(x + 5)^{2}-25+(y - 3)^{2}-9 = 35$, which simplifies to $(x + 5)^{2}+(y - 3)^{2}=35 + 25+9=69$. The center is $(-5,3)$.

Step2: Use center - radius form for second question

Given the center $(h,k)=(4,-3)$ and radius $r = 8$, the equation of the circle is $(x - 4)^{2}+(y+3)^{2}=r^{2}=(8)^{2}=64$.

Step3: Calculate radius for third question

The distance between the center $(h,k)=(3,-4)$ and the point $(x,y)=(6,-4)$ on the circle is the radius. Using the distance formula $d=\sqrt{(x - h)^{2}+(y - k)^{2}}$, we have $r=\sqrt{(6 - 3)^{2}+(-4+4)^{2}}=\sqrt{3^{2}} = 3$. The equation of the circle is $(x - 3)^{2}+(y + 4)^{2}=3^{2}=9$.

Answer:

1. C. $(-5,3)$
2. D. $(x - 4)^{2}+(y + 3)^{2}=64$
3. A. $(x - 3)^{2}+(y + 4)^{2}=9$