Question

what is the radius of a circle given by the equation x^2 + y^2 - 2x + 8y - 47 = 0? radius = units

Explanation:

Step1: Complete the square for x - terms

The x - terms are $x^{2}-2x$. Completing the square: $(x - 1)^{2}=x^{2}-2x + 1$, so $x^{2}-2x=(x - 1)^{2}-1$.

Step2: Complete the square for y - terms

The y - terms are $y^{2}+8y$. Completing the square: $(y + 4)^{2}=y^{2}+8y+16$, so $y^{2}+8y=(y + 4)^{2}-16$.

Step3: Rewrite the circle equation

Substitute the completed - square expressions into the original equation $x^{2}+y^{2}-2x + 8y-47 = 0$.
We get $(x - 1)^{2}-1+(y + 4)^{2}-16-47 = 0$.
Simplify it to $(x - 1)^{2}+(y + 4)^{2}=1 + 16+47$.

Step4: Calculate the radius

The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where r is the radius.
Since $(x - 1)^{2}+(y + 4)^{2}=64$, then $r^{2}=64$, and $r = 8$.

Answer:

8