Question

for the equation $x^{2}+y^{2}-6x - 2y-6 = 0$, do the following.
(a) find the center $(h,k)$ and radius $r$ of the circle.
(b) graph the circle.
(c) find the intercepts, if any.

(a) the center is $(3,1)$.
(type an ordered pair.)

the radius is $r = square$.

Explanation:

Step1: Rewrite the given equation in standard form

The general equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius. Given $x^{2}+y^{2}-6x - 2y-6 = 0$. Complete the square for $x$ and $y$ terms.
For the $x$ - terms: $x^{2}-6x=(x - 3)^{2}-9$. For the $y$ - terms: $y^{2}-2y=(y - 1)^{2}-1$.
So the equation becomes $(x - 3)^{2}-9+(y - 1)^{2}-1-6=0$, which simplifies to $(x - 3)^{2}+(y - 1)^{2}=16$.

Step2: Identify the radius

Since the standard - form of the circle is $(x - 3)^{2}+(y - 1)^{2}=16=(x - 3)^{2}+(y - 1)^{2}=4^{2}$, by comparing with $(x - h)^{2}+(y - k)^{2}=r^{2}$, we can see that $r = 4$.

Answer:

4