Question

what is the radius of the circle described by the equation $(x - 6)^{2}+(y - 4)^{2}-64 = 0$? a 64 b 16 c 32 d 8

Explanation:

Step1: Transform the equation to 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.
Given \((x - 6)^2+(y - 4)^2-64 = 0\), we can rewrite it as \((x - 6)^2+(y - 4)^2=64\).

Step2: Find the radius

Since \(r^2 = 64\), then \(r=\sqrt{64}\).
\(r = 8\)

Answer:

D. 8