Question

consider the following equation of a circle. (x^{2}+y^{2}+6x - 6y-18 = 0) step 2 of 3: find the radius, (r). answerhow to enter your answer (opens in new window) 2 points

Explanation:

Step1: Complete the square for x and y terms

We have the equation $x^{2}+y^{2}+6x - 6y-18 = 0$.
For the x - terms: $x^{2}+6x=(x + 3)^{2}-9$.
For the y - terms: $y^{2}-6y=(y - 3)^{2}-9$.
So the equation becomes $(x + 3)^{2}-9+(y - 3)^{2}-9-18=0$.

Step2: Rewrite the equation in standard form

$(x + 3)^{2}+(y - 3)^{2}-9-9-18=0$.
$(x + 3)^{2}+(y - 3)^{2}=18 + 9+9$.
$(x + 3)^{2}+(y - 3)^{2}=36$.
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.

Step3: Find the radius

Since $(x + 3)^{2}+(y - 3)^{2}=36=r^{2}$, then $r=\sqrt{36}=6$.

Answer:

6