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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 :

Option 1)
$x^{2}+y^{2}-4x^{2}y^{2}=0$

Option 2)
$x^{2}+y^{2}-2xy=0$

Option 3)
$x^{2}+y^{2}-16x^{2}y^{2}=0$

Option 4)
$x^{2}+y^{2}-2x^{2}y^{2}=0$

Answers (1)

best_answer

$x^{2}+y^{2}=1$

Let any point on circle be $\left ( cos\theta ,sin\theta \ \right )$

Eq.of tangent

$x\; cos\; \theta +y\; sin\; \theta =1$

$x-intercept=\left ( \frac{1}{cos\theta ,0} \right )$

$y-intercept=\left (0, \frac{1}{sin\theta } \right )$

Let mid-point of $PQ$ is $\left ( h,k \right )$

$h=\frac{1}{2\; cos\; \theta },\; k=\frac{1}{2\; sin\; \theta }$

$\Rightarrow cos\theta =\frac{1}{2h},\; sin\theta =\frac{1}{2k}$

$\therefore sin^{2}\theta +cos^{2}\theta =\frac{1}{4h^{2}}+\frac{1}{4k^{2}}=1$

$\therefore Locus\; \; \frac{1}{x^{2}}+\frac{1}{y^{2}}=4$

or

$x^{2}+y^{2}-4x^{2}y^{2}=0$

Option 1)

$x^{2}+y^{2}-4x^{2}y^{2}=0$

Option 2)

$x^{2}+y^{2}-2xy=0$

Option 3)

$x^{2}+y^{2}-16x^{2}y^{2}=0$

Option 4)

$x^{2}+y^{2}-2x^{2}y^{2}=0$

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