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  • The square of the tangent that can be drawn from any pt. on one circle to another circle is k times the product of the perpendicular distance of the point from the radical axis of the two circles,

The square of the tangent that can be drawn from any pt. on one circle to another circle is k times the product of the perpendicular distance of the point from the radical axis of the two circles, and the distance between their centres, where k =

 

Option: 1

1


Option: 2

3


Option: 3

4


Option: 4

2


Answers (1)

best_answer

We have to prove that \mathrm{PA}^2=2 \mathrm{PM} \cdot \mathrm{C}_1 \mathrm{C}_2

\mathrm{\begin{aligned} & P \equiv(a \cos \theta, a \sin \theta) \\ & P A=\sqrt{a^2-b^2+h^2-2 a h \cos \theta} \\ & \text { Radical axis: }-2 h x+h^2-a^2+b^2=0 \text { or } x=\frac{h^2-a^2+b^2}{2 h} \end{aligned}}

\mathrm{\begin{aligned} & P M=\frac{h^2-a^2+b^2}{2 h}-a \cos \theta=\frac{h^2-a^2+b^2-2 a h \cos \theta}{2 h} \\ & \left.P M \cdot C_1 C_2=\frac{h^2-a^2+b^2-2 a h \cos \theta}{2 h} \quad \text { (where } C_1 C_2=h\right) \quad=P A^2 / 2 \end{aligned}}

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Suraj Bhandari

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