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

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Solution

Equation of required circle
$x^{2} + y^{2} - 2 x + 4 y - 1 + λ (x^{2} + y^{2} + 4 x - 2 y - 5) = 0$
Apply condition of orthogonality with $x^{2} + y^{2} = 4$
$\frac{1 + 5 λ}{1 + λ} = - 4$
$\Rightarrow 1 + 5 λ = - 4 - 4 λ$
$\Rightarrow 9 λ = - 5 \Rightarrow λ = - \frac{5}{9}$
The length of tangent from origin $= \sqrt{S_{1}} = \sqrt{1} 6 = 4$

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A circle S=0 passes through points of intersection of circles x2+y2−2x+4y−1=0 and x2+y2+4x−2y−5=0 and cuts the circle x2+y2=4 orthogonally. Then length of tangent from origin on circle S=0 is

Equation of required circlex2+y2−2x+4y−1+λ(x2+y2+4x−2y−5)=0Apply condition of orthogonality with x2+y2=41+5λ1+λ=−4⇒1+5λ=−4−4λ⇒9λ=−5⇒λ=−59The length of tangent from origin =√S1=√16=4

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