Question

find the center and radius of the circle defined by the equation $x^{2}+y^{2}+4x - 2y - 20 = 0$. center: ( , ) radius:

Explanation:

Step1: Rewrite the equation

Group the $x$ - terms and $y$ - terms: $(x^{2}+4x)+(y^{2}-2y)=20$.

Step2: Complete the square for $x$

For $x^{2}+4x$, add $(\frac{4}{2})^{2}=4$ to both sides of the equation. So we have $(x^{2}+4x + 4)+(y^{2}-2y)=20 + 4$.

Step3: Complete the square for $y$

For $y^{2}-2y$, add $(\frac{-2}{2})^{2}=1$ to both sides of the equation. Then $(x^{2}+4x + 4)+(y^{2}-2y + 1)=20 + 4+1$.

Step4: Write in standard form

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. We get $(x + 2)^{2}+(y - 1)^{2}=25$, which is $(x-(-2))^{2}+(y - 1)^{2}=5^{2}$. So the center is $(-2,1)$ and the radius is $5$.

Answer:

center: $(-2,1)$
radius: $5$