JEE PYQ: Circle Question 26
Question 26 - 2019 (09 Apr Shift 1)
If a tangent to the circle $x^2 + y^2 = 1$ intersects the coordinate axes at distinct points $P$ and $Q$, then the locus of the mid-point of $PQ$ is:
(1) $x^2 + y^2 - 4x^2y^2 = 0$
(2) $x^2 + y^2 - 2xy = 0$
(3) $x^2 + y^2 - 16x^2y^2 = 0$
(4) $x^2 + y^2 - 2x^2y^2 = 0$
Show Answer
Answer: (1)
Solution
Tangent: $x\cos\theta + y\sin\theta = 1$. $P = (\sec\theta, 0)$, $Q = (0, \csc\theta)$. Mid-point $M(h,k)$: $h = \frac{1}{2\cos\theta}$, $k = \frac{1}{2\sin\theta}$. So $\frac{1}{4h^2} + \frac{1}{4k^2} = 1 \Rightarrow h^2 + k^2 = 4h^2k^2$.
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