Question

identifying the radius given an equation in general form
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

Group the x - terms and y - terms: $(x^{2}-2x)+(y^{2}+8y)=47$.
For the x - terms, $x^{2}-2x=(x - 1)^{2}-1$.

Step2: Complete the square for y - terms

For the y - terms, $y^{2}+8y=(y + 4)^{2}-16$.
So, $(x - 1)^{2}-1+(y + 4)^{2}-16 = 47$.

Step3: Rewrite the equation in standard form

$(x - 1)^{2}+(y + 4)^{2}=47 + 1+16$.
$(x - 1)^{2}+(y + 4)^{2}=64$.
The standard form of a circle equation is $(x - a)^{2}+(y - b)^{2}=r^{2}$, where r is the radius.

Step4: Find the radius

Since $r^{2}=64$, then $r = 8$.

Answer:

8