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  • The two circles which pass through the points (0, a), (0, −a) and touch the line y = mx + c will cut orthogonally, if <img alt="\mathrm{c}^2=\mathrm{a}^2\left(\mathrm{k}+\mathrm{m}^2\rig

The two circles which pass through the points (0, a), (0, −a) and touch the line y = mx + c will cut orthogonally, if \mathrm{c}^2=\mathrm{a}^2\left(\mathrm{k}+\mathrm{m}^2\right) , where k= 

Option: 1

1


Option: 2

2


Option: 3

-1


Option: 4

-2


Answers (1)

best_answer

Let the equation of circle be

        \mathrm{x^2+y^2+2 g x+2 f v+k=0} -----------(1)

It passes through (0, a) and (0, −a)

    \mathrm{\therefore \mathrm{a}^2+2 \mathrm{af}+\mathrm{k}=0} -----------(2)

and \mathrm{\therefore \mathrm{a}^2-2 \mathrm{af}+\mathrm{k}=0} --------(3)

From Eqns. (2) and (3),  \mathrm{f=0, k=-a^2}

and then the equation (1) of circle becomes,

\mathrm{x^2+y^2+2 g x-a^2=0} -------------(4)

Given that line y = mx + c touches this circle so that length of perpendicular from the centre (-g, 0) is equal to the radius 

  \mathrm{\sqrt{g^2+a^2}} .

\begin{aligned} & \Rightarrow \frac{-m g+c}{\sqrt{m^2+1}}=\sqrt{g^2+a^2} \\ & \text { or, } g^2+2 m c g+a^2\left(1+m^2\right)-c^2=0 \end{aligned}---------(5)

These circles cut orthogonally 

\mathrm{\begin{aligned} & 2\left[\mathrm{a}^2\left(1+\mathrm{m}^2\right)-\mathrm{c}^2\right]+2 \times 0 \times 0=-\mathrm{a}^2-\mathrm{a}^2 \quad\left(2 \mathrm{~g}_1 \mathrm{~g}_2+2 \mathrm{f}_1 \mathrm{f}_2=\mathrm{c}_1+\mathrm{c}_2\right) \\ & \Rightarrow \quad \mathrm{a}^2\left(2+\mathrm{m}^2\right)=\mathrm{c}^2 \end{aligned})

 

 

 

 

Posted by

Gautam harsolia

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