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Earlier discussion
1 older comment kept for context.
Victor Titus
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
\(-2a = -8; -2b = -2 \implies a = 4, b = 1\)
\(\therefore r^{2} - 4^{2} - 1^{2} = -1 \implies r^{2} = -1 + 16 + 1 = 16\)
\(r = \sqrt{16} = 4\)