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  • The circle  with centre <img alt="\mathrm{C_1\left(a_1, b_1\right)}" src="https://entranc

The circle \mathrm{ S_1} with centre \mathrm{C_1\left(a_1, b_1\right)} and radius \mathrm{r_1} touches externally the circle \mathrm{S_2} with centre \mathrm{C_2\left(a_2, b_2\right) and \: \: radius\: \: r_2}. If the tangent at their common point passes through the origin, then
 

Option: 1

\mathrm{\left(a_1^2+a_2^2\right)+\left(b_1^2+b_2^2\right)=r_1^2+r_2^2}

 


Option: 2

\mathrm{\left(a_1^2-a_2^2\right)+\left(b_1^2-b_2^2\right)=r_1^2-r_2^2}
 


Option: 3

\mathrm{\left(a_1^2-b_1^2\right)+\left(a_2^2-b_2^2\right)=r_1^2+r_2^2}
 


Option: 4

\mathrm{\left(a_1^2-b_1^2\right)+\left(a_2^2-b_2^2\right)=r_1^2-r_2^2}


Answers (1)

best_answer

The two circles are \mathrm{\left(x-a_1\right)^2+\left(y-b_1\right)^2=r_1^2}...........(i)
                            \mathrm{ \left(x-a_2\right)^2+\left(y-b_2\right)^2=r_2^2 } ..........(ii)

(i) - (ii) gives the equation to the common tangent

\mathrm{ 2 x\left(a_2-a_1\right)+2 y\left(b_2-b_1\right)+a_1^2+b_1^2-a_2^2-b_2^2-r_1^2+r_2^2=0 }

If this passes through the origin, then \mathrm{ \left(a_1^2-a_2^2\right)+\left(b_1^2-b_2^2\right)=r_1^2-r_2^2 }

Hence option 2 is correct.

Posted by

jitender.kumar

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