Question

determine the center and radius of the circle $x^{2}+y^{2}-18x + 4y-7 = 0$. enter the center as a point $(a,b)$. the center is . the radius is .

Explanation:

Step1: Complete the square for x - terms

We have $x^{2}-18x$. To complete the square, we take half of the coefficient of $x$ (-18), square it. Half of -18 is -9, and $(-9)^{2}=81$. So $x^{2}-18x=(x - 9)^{2}-81$.

Step2: Complete the square for y - terms

We have $y^{2}+4y$. Half of 4 is 2, and $2^{2}=4$. So $y^{2}+4y=(y + 2)^{2}-4$.

Step3: Rewrite the circle equation

The original equation $x^{2}+y^{2}-18x + 4y-7 = 0$ can be rewritten as $(x - 9)^{2}-81+(y + 2)^{2}-4-7=0$.

Step4: Simplify the equation

Combining like - terms, we get $(x - 9)^{2}+(y + 2)^{2}=81 + 4+7$.

Step5: Determine the center and radius

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. So, $(x - 9)^{2}+(y + 2)^{2}=92$. The center is $(9,-2)$ and the radius $r=\sqrt{92}=2\sqrt{23}$.

Answer:

The center is $(9,-2)$
The radius is $2\sqrt{23}$