Circle Ques 81
- Consider a family of circles passing through two fixed points $A(3,7)$ and $B(6,5)$. Show that 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. Find the coordinates of this point.
(1993, 5M)
Show Answer
Answer:
Correct Answer: 81.$x=2$ and $y=23 / 3$
Solution:
Formula:
- The equation of the circle on the line joining the points $A(3,7)$ and $B(6,5)$ as diameter is
$$ (x-3)(x-6)+(y-7)(y-5)=0 $$
and the equation of the line joining the points $A(3,7)$ and $B(6,5)$ is $y-7=\frac{7-5}{3-6}(x-3)$
$\Rightarrow \quad 2 x+3 y-27=0$
Now, the equation of family of circles passing through the point of intersection of Eqs. (i) and (ii) is
$$ S+\lambda P=0 $$
$\Rightarrow(x-3)(x-6)+(y-7)(y-5)+\lambda(2 x+3 y-27)=0$
$\Rightarrow x^{2}-6 x-3 x+18+y^{2}-5 y-7 y+35$ $+2 \lambda x+3 \lambda y-27 \lambda=0$
$\Rightarrow S _1 \equiv x^{2}+y^{2}+x(2 \lambda-9)+y(3 \lambda-12)$
$$ +(53-27 \lambda)=0 $$
Again, the circle,which cuts the members of family of circles, is
$$ S _2 \equiv x^{2}+y^{2}-4 x-6 y-3=0 $$
and the equation of common chord to circles $S _1$ and $S _2$ is
$$ \begin{array}{rlrl} & & S _1-S _2 & =0 \\ & \Rightarrow & x(2 \lambda-9+4)+y(3 \lambda-12+6)+(53-27 \lambda+3) & =0 \\ \Rightarrow & 2 \lambda x-5 x+3 \lambda y-6 y+56-27 \lambda & =0 \\ \Rightarrow & (-5 x-6 y+56)+\lambda(2 x+3 y-27) & =0 \end{array} $$
which represents equations of two straight lines passing through the fixed point whose coordinates are obtained by solving the two equations
$5 x+6 y-56=0$ and $2 x+3 y-27=0$,
we get $x=2$ and $y=23 / 3$
BETA
