Question

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. $(6,1)$ d. $(0,1)$ what is the center of the circle with the equation $(x + 4)^2+(y - 3)^2=16$? a. $(-4,3)$ b. $(4,3)$ c. $(4,-3)$ d. $(-4,-3)$ if the equation of a circle is $x^2 + y^2=144$, what is the radius of the circle? a. 144 b. 36 c. 12 d. 72

Explanation:

Step1: Recall circle - equation formula

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 and $r$ is the radius.

First question:

We have the circle equation $(x + 3)^2+(y - 1)^2 = 9$, which is in the form $(x-(-3))^2+(y - 1)^2=3^2$. To check which point lies on the circle, we substitute the $x$ and $y$ - values of each point into the equation.
For point $(-3,1)$:
Substitute $x=-3$ and $y = 1$ into the equation $(x + 3)^2+(y - 1)^2$:
\[(-3 + 3)^2+(1 - 1)^2=0^2+0^2=0
eq9\]
For point $(3,-1)$:
Substitute $x = 3$ and $y=-1$ into the equation:
\[(3 + 3)^2+(-1 - 1)^2=6^2+(-2)^2=36 + 4=40
eq9\]
For point $(6,1)$:
Substitute $x = 6$ and $y = 1$ into the equation:
\[(6 + 3)^2+(1 - 1)^2=9^2+0^2=81
eq9\]
For point $(0,1)$:
Substitute $x = 0$ and $y = 1$ into the equation:
\[(0 + 3)^2+(1 - 1)^2=3^2+0^2=9\]

Second question:

The circle equation is $(x + 4)^2+(y - 3)^2 = 16$, which is in the form $(x-(-4))^2+(y - 3)^2=4^2$. The center of the circle in the standard - form $(x - a)^2+(y - b)^2=r^2$ is $(a,b)$. So, the center of the circle $(x + 4)^2+(y - 3)^2 = 16$ is $(-4,3)$.

Third question:

The circle equation is $x^2+y^2 = 144$, which is in the form $(x - 0)^2+(y - 0)^2=12^2$. In the standard - form of the circle equation $(x - a)^2+(y - b)^2=r^2$, comparing $x^2+y^2 = 144$ with $(x - 0)^2+(y - 0)^2=r^2$, we have $r^2=144$, so $r = 12$.

Answer:

1. d. $(0,1)$
2. a. $(-4,3)$
3. c. $12$