Find the radius of the circle \(x^{2} + y^{2} - 8x - 2y + 1

Explanation

Given the equation of the circle \(x^{2} + y^{2} - 8x - 2y + 1 = 0\).

The equation of a circle is given as \((x - a)^{2} + (y - b)^{2} = r^{2}\)

Expanding, we have \(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2} \equiv x^{2} - 2ax + y^{2} - 2by = r^{2} - a^{2} - b^{2}\)

Comparing the RHS of the equation above with the equation rewritten as \(x^{2} + y^{2} - 8x - 2y = -1\), we have

\(-2a = -8; -2b = -2 \implies a = 4, b = 1\)

\(\therefore r^{2} - 4^{2} - 1^{2} = -1 \implies r^{2} = -1 + 16 + 1 = 16\)

\(r = \sqrt{16} = 4\)