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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 point P and Q. then the locus of the mid- point of PQ is

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
Equation of given circles is `x^(2)+y^(2)=1`, then equation of tangent at the point `(costheta, sintheta)` on the given circle is
`xcostheta+y sintheta=1" "...(i)`
[`therefore` Equation of tangent at the point `P(costheta, sintheta)` to the circle `x^(2) +y^(2)=r^(2) is x cos theta+ y sintheta=r`]
Now, the point of intersection with coordinates axes are `P(sectheta, 0) and Q(0,cosec theta)`
`therefore` Mid-point of line joining points P and Q is
`M((sectheta)/(2),(cosectheta)/(2))=(h,k)` (let)
`So, costheta=(1)/(2h)and sintheta=(1)/(2k)`
`therefore sin^(2)theta+cos^(2)theta=1`
`therefore (1)/(4h^(2))+(1)/(4k^(2))=1rArr(1)/(h^(2))+(1)/(k^(2))=4`
Now, locus of mid-point M is
`(1)/(x^(2))+(1)/(y^(2))=4`
`rArr x^(2)+y^(2)-4x^(2)y^(2)=0`
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