The circle x^{2}+y^{2}-2kx-2ky+k^{2}=0 touches the x-axis at P and y-axis at Q. What is PQ equal to?

Explanation

The given circle can be rewritten as (x-k)^2 + (y-k)^2 = k^2, which has its center at (k, k) and radius k. It touches the x-axis at P(k, 0) and the y-axis at Q(0, k). The distance PQ is \sqrt{(k-0)^2 + (0-k)^2} = \sqrt{k^2+k^2} = \sqrt{2}k.

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