1.

Find the circum-centre and the circum-radius of a triangle ABC formed by the vertices A(2,-2) , B (-1,1) and C(3,1).

Answer» Let S(x,y) be the circum-centre of `Delta`ABC .
`therefore SA^(2) = SB^(2) = SC^(2)`
Consider `SA^(2) = SB^(2)`
`implies (x-2)^(2) + (y+2)^(2) = (x +1)^(2) + (y-1)^(2)`
`x^(2) - 4x + 4 + y^(2) + 4y + 4 = x^(2) + 2x + 1 + y -2y + 1`
`-4x + 4y + 8 = 2x -2y + 2`
`6x - 6y - 6 = 0`
`x - y - 1= 0" " (1)`
`SB^(2) = SC^(2) `
`implies (x+1)^(2) + (y-1)^(2) = (x-3)^(2) + (y-1)^(2)`
`x^(2) + 2x + 1 + y^(2) - 2y + 1 = x^(2) - 6x + 9 + y - 2y +1`
`2x - 2y + 2 = -6x - 2y + 10`
`8x - 8 = 0`
`implies x -1`.
Substituting x = 1 in Eq. (1) , we get y = 0 .
`therefore` The required circum-centre of `DeltaABC` is (1,0).
Circum-radius , SA = `sqrt((2-1)^(2) + (-2-0)^(2)) = sqrt5` units.


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