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  • Consider a family of circles passing through two fixed points A(3, 7) and B(6, 5). The chords in which the circle  <img alt="\mathrm{x^2+y^2-4 x-6 y-3=0}" src="https://entrancecorner.oncodecog

Consider a family of circles passing through two fixed points A(3, 7) and B(6, 5). The chords in which the circle  \mathrm{x^2+y^2-4 x-6 y-3=0}cuts the members of the family are concurrent at a point. Find the co-ordinates of this point.

 

Option: 1

\mathrm{\left(2, \frac{23}{3}\right)}


Option: 2

\mathrm{(-2,-3)}


Option: 3

\mathrm{(2,4)}


Option: 4

\mathrm{\left(2, \frac{-23}{4}\right)}


Answers (1)

best_answer

The equation of a circle through the given points A(3, 7) and B(6, 5) is \mathrm{S}+\lambda P=0 where

S = 0 is equation of circle on AB as diameter and P = 0 is the equation of line AB

\mathrm{\begin{aligned} & (x-3)(x-6)+(y-7)(y-5)+\lambda(2 x+3 y-27)=0 \\ & \mathrm{~S}=\mathrm{x}^2+\mathrm{y}^2+(2 \lambda-9) \mathrm{x}+(3 \lambda-12) \mathrm{y}+53-27 \lambda=0 \\ & \mathrm{~S}_1=\mathrm{x}^2+\mathrm{y}^2-4 \mathrm{x}-6 \mathrm{y}-3=0 \end{aligned}}

Common chord is given by \mathrm{\mathrm{S}_1-\mathrm{S}_2=0}

\mathrm{(2 \lambda-5) x+(3 \lambda-6) y+(56-27 \lambda)=0 \text { or } \quad(-5 x-6 y+56)+\lambda(2 x+3 y-27)=0}

It is of the form \mathrm{P+\lambda Q=0} 

which represents a family of lines passing through the intersection of 

P = 0 and Q = 0 i.e. the point .

 

 

Posted by

Gaurav

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