Question

# A circle $$S = 0$$ passes through points of intersection of circles $$x^2 + y^2 - 2x + 4y -1 = 0$$ and $$x^2 + y^2 + 4x- 2y - 5 = 0$$ and cuts the circle $$x^2 + y^2 = 4$$ orthogonally. Then length of tangent from origin on circle $$S = 0$$ is

Solution

## Equation of required circle$$x^2 + y^2 -2x + 4y -1 + \lambda (x^2 + y^2 + 4x -2y- 5) = 0$$Apply condition of orthogonality with $$x^2 + y^2 = 4$$$$\dfrac {1+5\lambda}{1+\lambda}=-4$$$$\Rightarrow 1+5\lambda=-4-4\lambda$$$$\Rightarrow 9\lambda =-5\Rightarrow \lambda=-\dfrac {5}{9}$$The length of tangent from origin $$=\sqrt {S_1}=\sqrt 16=4$$Maths

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