Question

1. determine the center and radius of the following circle $x^{2}+y^{2}=49$.
2. determine the center and radius of the following circle $(x - 2)^{2}+(y + 3)^{2}=25$.
3. determine the center and radius of the following circle $x^{2}+y^{2}+6x - 8y - 11=0$.

Explanation:

Step1: Recall circle - standard form

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

Step2: Solve for the first circle $x^{2}+y^{2}=49$

Since $x^{2}+y^{2}=(x - 0)^2+(y - 0)^2 = 49=7^{2}$, the center is $(0,0)$ and the radius $r = 7$.

Step3: Solve for the second circle $(x - 2)^2+(y + 3)^2=25$

Comparing with the standard - form $(x - a)^2+(y - b)^2=r^2$, we have $a = 2$, $b=-3$, and $r = 5$. So the center is $(2,-3)$ and the radius is $5$.

Step4: Rewrite the third circle $x^{2}+y^{2}+6x - 8y-11 = 0$ in standard form

Complete the square for $x$ and $y$ terms:

$$
LATEXBLOCK0
$$

Comparing with $(x - a)^2+(y - b)^2=r^2$, we get $a=-3$, $b = 4$, and $r = 6$. So the center is $(-3,4)$ and the radius is $6$.

Answer:

1. Center: $(0,0)$

Radius: $7$

1. Center: $(2,-3)$

Radius: $5$

1. Center: $(-3,4)$

Radius: $6$