1.

If the circle x^(2) + y^(2) + 2x + 3y + 1 = 0 cuts another circle x^(2) + y^(2) + 4x + 3y + 2 = 0 in A and B, then the equation of the circle with AB as a diamter is

Answer»

`x^(2) + y^(2) + x + 3y + 3 = 0`
`2X^(2) + 2y^(2) + 2x + 6Y + 1 = 0`
`x^(2) + y^(2) + x + 6y + 1 = 0`
`2x^(2) + 2y^(2) + x + 3y + 1 = 0`

Solution :Given circle are,
`S = x^(2) + y^(2) + 2x + 3y + 1 = 0` and
`S^(1) -= x^(2) + y^(2) + 4x + 3y + 2 = 0`
`implies` EQUATION of radical axis is `S - S^(1) = 0`
-2x - 1 = 0
2x + 1 = 0
Equation of required circle is
`implies (x^(2) + y^(2) + 2x + 3y + 1) + lambda (2x + 1) = 0` Centre, `C = (-1 - lambda, (-3)/(2))` LIES on equation (1) then
`implies - 2 - 2 lambda + 1 = 0`
`implies 2 lambda = - 1`
`implies lambda = (-1)/(2)`
Equation (2) `implies x^(2) + y^(2) + 2x + 3y + 1 - (1)/(2) (2x + 1) = 0`
`implies 2x^(2) + 2y^(2) + 4x + 6y + 2 - 2x - 1 = 0`
`:. 2x^(2) + 2y^(2) + 2x + 6y + 1 = `, is required circle


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