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The circle x^2+y^2-4x-4y+4=0 is inscribe...

The circle `x^2+y^2-4x-4y+4=0` is inscribed in a variable triangle `O A Bdot` Sides `O A` and `O B` lie along the x- and y-axis, respectively, where `O` is the origin. Find the locus of the midpoint of side `A Bdot`

Text Solution

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Given equation of circle is `(x-2)^(2)+(y-2)^(2)=2^(2)`
It touches both the axes

Let the midpoint of AB be P(h,k)
`:. A-=(2h,0)` and `B-=(0,2k)`
So, the equation of AB is `(x)/(2h)+(y)/(2k)=1` ltbgt Since this line touches the given circle, we have
`(|(2)/(2h)+(2)/(2k)-1|)/sqrt((1)/(4h^(2))+(1)/(4k^(2)))=2`
On simplifying, we get locus of point P as `x+y-xy+sqrt(x^(2)+y^(2))=0`.
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