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)

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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