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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 |
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Answer» `x^(2) + y^(2) + x + 3y + 3 = 0` `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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