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If a tangent to the circle x^(2)+y^(2)=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 :

A

`x^(2)+y^(2)-2xy=0`

B

`x^(2)+y^(2)-16x^(2)y^(2)=0`

C

`x^(2)+y^(2)-2x^(2)y^(2)=0`

D

`x^(2)+y^(2)-4x^(2)y^(2)=0`

Text Solution

Verified by Experts

The correct Answer is:
D
`rArr` Mid-point `A,B=P=(h/2,k/2)`
`rArr h=2x,k=2y`
Now line as: `x/h+y/x=1` is layout
`rArr (|0/h + 0/k -1|)/(sqrt((1)/(h^(2))+(1)/(k^(2))))=1 rArr (1)/(h^(2))+(1)/(x^(2))=1`
So, locus: `(1)/(4x^(2))+(1)/(4y^(2))=1 rArr x^(2)+y^(2)=4x^(2)y^(2) rArr x^(2)+y^(2)-4x^(2)y^(2)=0`
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