Question

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

• A. $(2,1)$
• B. $\left(2,\frac{23}{3}\right)$
• C. $(3,2)$
• D. $(4,3)$

Answer 1

Answer: (B)

The equation of the line passing through the points $A(3,7)$ and $B(6,5)$ is
$y-7=-\frac{2}{3}(x-3)$
or $2x+3y-27=0$
Also, the equation of the circle with $A$ and $B$ as the endpoints of diameter is
$(x-3)(x-6)+(y-7)(y-5)=0$
Now, the equation of the family of circles through $A$ and $B$ is
$(x-3)(x-6)+(y-7)(y-5)+\lambda (2x+3y-27)=0$
The equation of the common chord of (i)
and $x^2+y^2-4x-6y-3=0$ is the radical axis, which is
This is the family of lines which passes through the point of intersection of
$-5x-6y+56=0$ and $2x+3y-27=0$,
i.e., $(2,23/3)$.