Question

1. what is the center of a circle with the equation x² + y² + 10x - 6y - 35?

a. (5,3)
b. (5, - 3)
c. (-5,3)
d. (-5, - 3)

1. 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)

1. which of the following points lies on the circle with equation (x + 3)² + (y - 1)² = 9?

a. (-3,1)
b. (3, - 1)
c. (0,1)
d. (0,1)

Explanation:

5.

Step1: Rewrite the equation in standard form

The general equation of a circle is $x^{2}+y^{2}+Dx + Ey+F = 0$, and its standard - form is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where the center is $(a,b)$. Given $x^{2}+y^{2}+10x - 6y-15 = 0$. Complete the square for $x$ and $y$ terms. For the $x$ - terms: $x^{2}+10x=(x + 5)^{2}-25$. For the $y$ - terms: $y^{2}-6y=(y - 3)^{2}-9$. Then the equation becomes $(x + 5)^{2}-25+(y - 3)^{2}-9-15 = 0$, which simplifies to $(x + 5)^{2}+(y - 3)^{2}=49$. The center of the circle is $(-5,3)$.

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where $(a,b)$ is the center of the circle. Given the equation $(x + 8)^{2}+(y - 6)^{2}=36$, comparing it with the standard form, we have $x-(-8)$ and $y - 6$. So the center of the circle is $(-8,6)$.

The equation of the circle is $(x + 3)^{2}+(y - 1)^{2}=9$. A point $(x_0,y_0)$ lies on the circle if it satisfies the circle's equation.

Step1: Check option a

For the point $(-3,1)$, substitute $x=-3$ and $y = 1$ into the left - hand side of the equation: $(-3 + 3)^{2}+(1 - 1)^{2}=0+0 = 0
eq9$.

Step2: Check option b

For the point $(3,-1)$, substitute $x = 3$ and $y=-1$: $(3 + 3)^{2}+(-1 - 1)^{2}=36 + 4=40
eq9$.

Step3: Check option c

For the point $(0,1)$, substitute $x = 0$ and $y = 1$: $(0 + 3)^{2}+(1 - 1)^{2}=9+0 = 9$.

Step4: Check option d

For the point $(0,1)$ (repeated option), we already know from step 3 that when $x = 0$ and $y = 1$, $(x + 3)^{2}+(y - 1)^{2}=9$.

Answer:

c. $(-5,3)$

6.