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Consider a family of circles passing through two fixed points $A(3,7)$ & $B(6,5)$ . Find the point of concurrency of the chords in which the circle $x^{2}+y^{2}-4x-6y-3=0$ cuts the members of the family :
Option: 1
$\left ( \frac{11}{17},\frac{3}{7} \right )$

Option: 2
$\left ( 2,\frac{23}{3}\right )$

Option: 3
$\left (-4,3\right )$

Option: 4
chords are not concurrent

Answers (1)

best_answer

Equation of a circle in diametric form -

$\left ( x-x_{1} \right )\left ( x-x_{2} \right )+\left ( y-y_{1} \right )\left ( y-y_{2} \right )= 0$

- wherein

Where $A\left ( x_{1},y_{1} \right )\, and \:B \left ( x_{2},y_{2} \right )$ are the two diametric ends.

Family of circle -

$S+KL= 0$

- wherein

Equation of the family of circles passing through point of intersection $S=0 \, and\, line\, L=0$ .

Equation of the family of circles passing through $A(3,7)$ and $B(6,5)$ is

$(x-3)(x-6)+(y-7)(y-5)+\lambda (2x+3y-27)=0$

Equation of given circle is $x^{2}+y^{2}-4x-6y-3=0$

$\Rightarrow$ Equation of common chord is :- $S_{1}-S_{2}=0$

$\Rightarrow (2\lambda -5)x+(3\lambda -6)y+(56-27\lambda )=0$

$\Rightarrow \lambda(2x+3y-27)-(5x+6y-56)=0$

$\Rightarrow$ This represents family of lies passing through the point of intersection of $2x+3y-27=0$ & $5x+6y-56=0$

$\Rightarrow$ fixed point = $\left ( 2,\frac{23}{3} \right )$

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