SATHEE: Circle 1 Question 25

Circle 1 Question 25

25. Let $P$ be a point on the circle $S$ with both coordinates being positive. Let the tangent to $S$ at $P$ intersect the coordinate axes at the points $M$ and $N$ . Then, the mid-point of the line segment $M N$ must lie on the curve

(a) $(x + y)^{2} = 3 x y$

(b) $x^{2 / 3} + y^{2 / 3} = 2^{4 / 3}$

(c) $x^{2} + y^{2} = 2 x y$

(d) $x^{2} + y^{2} = x^{2} y^{2}$

Show Answer

Solution:

We have,

$x^{2} + y^{2} = 4$

Let $P (2 \cos θ, 2 \sin θ)$ be a point on a circle.

$∴$ Tangent at $P$ is

$2 \cos θ x + 2 \sin θ y = 4$

$\Rightarrow x \cos θ + y \sin θ = 2$

$∴$ The coordinates at $M \frac{2}{\cos θ}, 0$ and $N 0, \frac{2}{\sin θ}$

Let $(h, k)$ is mid-point of $M N$ $∴ h = \frac{1}{\cos θ}$ and $k = \frac{1}{\sin θ}$

$\Rightarrow \cos θ = \frac{1}{h}$ and $\sin θ = \frac{1}{k}$

$\Rightarrow \cos^{2} θ + \sin^{2} θ = \frac{1}{h^{2}} + \frac{1}{k^{2}} \Rightarrow 1 = \frac{h^{2} + k^{2}}{h^{2} \cdot k^{2}}$

$\Rightarrow h^{2} + k^{2} = h^{2} k^{2}$

$∴$ Mid-point of $M N$ lie on the curve $x^{2} + y^{2} = x^{2} y^{2}$

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