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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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Answer» 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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