For circle \( C_{1} \):
\[
x^{2} + y^{2} - 4x - 2y + 1 = 0
\]
We can complete the square for \( x \) and \( y \).
For \( x \):
\[
x^{2} - 4x = (x - 2)^{2} - 4
\]
For \( y \):
\[
y^{2} - 2y = (y - 1)^{2} - 1
\]
Substituting these back into the equation:
\[
(x
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