Question

Consider a family of circles passing through two fixed points $A(3,7)$ and $B(6,5)$. The chords in which the circle $x^{2}+y^{2}-4 x-6 y-3=0$ 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 $ 2 x+3 y-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
$(x-3)(x-6)+(y-7)(y-5)+\lambda(2 x+3 y-27)=0$
The equation of the common chord of (i)
and $x^{2}+y^{2}-4 x-6 y-3=0$ is the radical axis, which is
This is the family of lines which passes through the point of intersection of
$-5 x-6 y+56=0$ and $2 x+3 y-27=0$,
i.e., $(2,23 / 3)$.