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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 x2 + y2 − 4x − 6y − 3 = 0 cuts the members of the family are

Consider a family of circles passing through two fixed points A(3, 7) and B(6, 5). The chords in which the circle x2 + y2 − 4x − 6y − 3 = 0 cuts the members of the family are concurrent at a point. Find the coordinates of this point.

 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

\mathrm{(2,7)}


Answers (1)

best_answer

Equation of chord AB, is 

\mathrm{\begin{aligned} y-7=\frac{5-7}{6-3}(x-3) \\ \Rightarrow \quad 2 x+3 y & =27 \end{aligned}} -------(1)

The equation of family of circles passing through A and B can be written as

\mathrm{\begin{aligned} & (x-3)(x-6)+(y-7)(y-5)+\lambda(2 x+3 y-27)=0 \\ & \text { i.e., } x^2+y^2-9 x-12 y+53+\lambda(2 x+3 y-27)=0 \end{aligned}} ----------(2)

Also, the given circle is

\mathrm{x^2+y^2-4 x-6 y-3=0}----------(3)

The common chord of circles (2) and (3) is

It is obvious that all the lines of this family will pass through the point of intersection of 

\mathrm{5 x+64-56=0 \text { and } 2 x+3 y-27=0}

Solving these, the point of concurrency is  \mathrm{\left(2, \frac{23}{3}\right)}
 

 

 

Posted by

Ritika Kankaria

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