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Tangents are drawn to the circle \mathrm{x^2+y^2=10}  at the points where it is met by the circle \mathrm{x^2+y^2-6 x-4 y+10=0}  then the point of intersection of these tangents is
 

Option: 1

\left(\frac{10}{3}, \frac{5}{3}\right)


Option: 2

 (2,3)
 


Option: 3

 (4,5)
 


Option: 4

 (3,2)
 


Answers (1)

best_answer

Equation of common chord is 

\mathrm{S}-\mathrm{S}^{\prime}=0 \Rightarrow 3 \mathrm{x}+2 \mathrm{y}-10=0

The chord of contact of required point w.r.t. circle \mathrm{x^2+y^2=10}  is also the common chord of two circles. Chord of contact of \mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)\text{ is }\mathrm{xx}_1+\mathrm{yy}_1-10=0
\mathrm{\frac{x_1}{3}=\frac{y_1}{2}=\frac{-10}{-10} \Rightarrow\left(x_1, y_1\right)=(3,2) }

Posted by

Rakesh

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