Question

Consider a family of circles passing through the points $(3,7)$ and $(6,5)$. The chords in which the circle $x^2+y^2-4x-6y-3=0$ cuts the family of circles are concurrent at the point

• A. $(-2,\frac{23}{3})$
• B. $(2,\frac{23}{3})$
• C. $(2,-\frac{23}{3})$
• D. $(-2,-\frac{23}{3})$

Answer 1

The correct option is B $(2,\frac{23}{3})$

The line through $(3,7)$ and $(6,5)$ is $\frac{y-7}{x-3}=-\frac{2}{3}$
or $2x+3y-27=0$
The family of circles is $(x-3)(x-6)+(y-7)(y-5)+\lambda (2x+3y-27)=0$
or $x^2+y^2-9x-12y+53+\lambda (2x+3y-27)=0$
or $x^2+y^2+(2\lambda -9)x+(3\lambda -12)y+53-27\lambda =0$
The chord along which this cuts the circle
$x^2+y^2-4x-6y-3=0$ is $S-S^{\prime }=0$
$(2\lambda -5)x+(3\lambda -6)y+56-27\lambda =0$
or $\lambda (2x+3y-27)-(5x+6y-56)=0$
These lines are concurrent at the point given by
$2x+3y-27=0$ and $5x+6y-56=0$
Using Rule of Cross multiplication, we get
$\therefore \frac{x}{3(-56)+6\times 27}=\frac{y}{-5\times 27+2\times 56}=\frac{1}{2\times 6-5\times 3}$
Solving $\frac{x}{3(-56)+6\times 27}=\frac{1}{2\times 6-5\times 3}$ and $\frac{y}{-5\times 27+2\times 56}=\frac{1}{2\times 6-5\times 3}$
we get $x=2,y=\frac{23}{3}$