What is the radius of the circle described by the equation (×-6)tiny 2 + (y -4)tiny 2 -64 = 0?

1 answer

To find the radius of the circle described by the equation, we start with the given equation:

\[ (x - 6)^2 + (y - 4)^2 - 64 = 0 \]

We can rearrange this equation to isolate the circle on one side:

\[ (x - 6)^2 + (y - 4)^2 = 64 \]

This equation is now in the standard form of a circle, which is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

where \((h, k)\) is the center of the circle, and \(r\) is the radius.

From the equation \((x - 6)^2 + (y - 4)^2 = 64\), we can identify:

  • \(h = 6\)
  • \(k = 4\)
  • \(r^2 = 64\)

To find the radius \(r\), we take the square root of \(r^2\):

\[ r = \sqrt{64} = 8 \]

Thus, the radius of the circle is:

\[ \boxed{8} \]