Question

find the center and radius of the circle. x^{2}+y^{2}-4x - 18y-7 = 0 the center is. (type an ordered pair.)

Explanation:

Step1: Rearrange and complete the square for x - terms

Group the x - terms and y - terms: $(x^{2}-4x)+(y^{2}-18y)=7$. Completing the square for the x - terms: $x^{2}-4x=(x - 2)^{2}-4$.

Step2: Complete the square for y - terms

Completing the square for the y - terms: $y^{2}-18y=(y - 9)^{2}-81$. So, $(x - 2)^{2}-4+(y - 9)^{2}-81 = 7$.

Step3: Rewrite the equation in standard form

Rearrange the equation to the standard form of a circle $(x - a)^{2}+(y - b)^{2}=r^{2}$: $(x - 2)^{2}+(y - 9)^{2}=7 + 4+81$. Then $(x - 2)^{2}+(y - 9)^{2}=92$.

Answer:

Center: $(2,9)$, Radius: $\sqrt{92}$