NDA 2 2025 Maths, question 64

Topic: Analytical geometry

Q64

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?

Solution.

Answer: (a) $\sqrt{2}k$

Completing squares: $(x - k)^{2} + (y - k)^{2} = k^{2}$ , centre $(k, k)$ , radius $k$ .
It touches the axes at $P(k, 0)$ and $Q(0, k)$ .
$PQ = \sqrt{k^{2} + k^{2}} = \sqrt{2}k$ .

Trap: Strictly $PQ = \sqrt{2}|k|$ ; the option takes $k > 0$ .