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The chord of contact of tangents from three points \mathrm{P, Q, R} to the circle \mathrm{x^2+y^2=c^2} are concurrent, then \mathrm{P, Q, R}
 

Option: 1

 form a triangle
 


Option: 2

 are concyclic
 


Option: 3

 are collinear
 


Option: 4

none of these


Answers (1)

best_answer

 Let \mathrm{P \equiv\left(x_1, y_1\right), Q \equiv\left(x_2, y_2\right), R \equiv\left(x_3, y_3\right)}
Equation of chords of contact of tangents from \mathrm{P, Q, R} to circle \mathrm{x^2+y^2=c^2} are

\begin{aligned} & x x_1+y y_1-a^2=0 & \dots (i)\\ & x x_2+y y_2-a^2=0 & \dots (ii) \\ & x x_3+y y_3-a^2=0 & \dots (iii) \end{aligned}

Since lines (i), (ii) and (iii) are concurrent,

\therefore\left|\begin{array}{lll} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array}\right|=0 \Rightarrow \text{Points P, Q, R are collinear.}
 

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