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Consider the family ol circles x^2+y^2=...

Consider the family ol circles `x^2+y^2=r^2, 2 < r < 5` . If in the first quadrant, the common tangnet to a circle of this family and the ellipse `4x^2 +25y^2=100` meets the co-ordinate axes at A and B, then find the equation of the locus of the mid-point of AB.

A

`4x^(2)+25y^(2)=4x^(2)y^(2)`

B

`4x^(2)+25y^(2)=4`

C

`4x^(2)+25y^(2)=9x^(2)y^(2)`

D

`4x^(2)+25y^(2)=9`

Text Solution

Verified by Experts

The correct Answer is:
A
Equation of any tangent to circle `x^(2)+y^(2)=r^(2)` is
`xcostheta + y sintheta = r" "...(i)`
Suppose Eq. (i) is tangent to `4x^(2) + 25y^(2)=100`
or when `(x^(2))/(25)+(y^(2))/(4)=1 " at"(x_(1),y_(2))`
Then, Eq, (i) and `("xx"_(1))/(25)+(yy_(1))/(4)=1` are identical
`therefore (x_(1)//25)/(costheta)=((y_(1))/(4))/(sintheta)=(1)/(r)`
`rArr x_(1)(25costheta)/(r),y_(1)=(4sintheta)/(r)`
The line (i) meet the coordinates axes in A `(rsectheta, 0) and beta(0,r cosec theta)`. LEt (h,k) be mid-point of AB.
Then , " " `h=(rsectheta)/(2)and k=(rcosectheta)/(2)`
Therefore, `2h=(r)/(costheta)and 2k=(r)/(sintheta)`
`therefore x_(1)=(25)/(2h)and y_(1)=(4)/(2k)`
As `(x_(1), y_(1)) " lies on the ellipse " (x^(2))/(25)+(y^(2))/(4)=1` we get
`(1)/(25)((625)/(4h^(2)))+(1)/(4)((4)/(k^(2)))=1`
`rArr (25)/(4h^(2))+(1)/(h^(2))=1`
or `25k^(2)+4h^(2)=4h^(2)k^(2)`
Therefore, required locus is `4x^(2)+25y^(2)=4x^(2)y^(2)`
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