Question

question 3 of 20
what is the radius of the circle shown below?
$x^{2}+y^{2}-12x - 6y+9 = 0$

Explanation:

Step1: Rewrite the equation in standard form

The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius.
Complete the square for $x$ and $y$ terms in $x^{2}+y^{2}-12x - 6y+9 = 0$.
For the $x$ - terms: $x^{2}-12x=(x - 6)^{2}-36$.
For the $y$ - terms: $y^{2}-6y=(y - 3)^{2}-9$.
So the equation becomes $(x - 6)^{2}-36+(y - 3)^{2}-9 + 9=0$.

Step2: Simplify the equation

$(x - 6)^{2}+(y - 3)^{2}=36$.

Step3: Identify the radius

Comparing with $(x - a)^2+(y - b)^2=r^2$, we have $r^{2}=36$, so $r = 6$.

Answer:

6