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Find the equation of a circle, the equation of whose two diameters are `x + y = 6 and 3x +4y =16` and radius is 10.

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We know that the two diameters of a circle inersect at the centre of the circle
`:.` The point of intersection `(8,-2)` of `x +y = 6 and 3x +4y =16` will be the centre of the circle.
`:.` Equation of circle `(x-8)^(2)+(y+2)^(2) =10^(2)`
`rArr x^(2)-16x+64+y^(2)+4y+4=100`
`rArrx^(2)+y^(2)-16x+4y-32=0`.
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