Question

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

Explanation:

Step1: Rearrange and complete the square for \(x\) terms

Group \(x\) - terms and \(y\) - terms: \((x^{2}-2x)+(y^{2}+8y)=47\).
For \(x\) - terms: \(x^{2}-2x=(x - 1)^{2}-1\) (using \((a - b)^{2}=a^{2}-2ab + b^{2}\), here \(a=x\), \(b = 1\)).

Step2: Complete the square for \(y\) terms

For \(y\) - terms: \(y^{2}+8y=(y + 4)^{2}-16\) (using \((a + b)^{2}=a^{2}+2ab + b^{2}\), here \(a=y\), \(b = 4\)).

Step3: Substitute back into the equation

Substitute into \((x^{2}-2x)+(y^{2}+8y)=47\):
\((x - 1)^{2}-1+(y + 4)^{2}-16=47\).

Step4: Simplify to the standard circle equation

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

Answer:

\(8\)