Question

Let a variable point $A$ lie on a line $L=0$. Tangents are drawn to the circle $x^2+y^2=9$ through the variable point $A$ meet the circle at $B$ and $D$. A point $C$ lies on the other side of the circle and also on the line $L=0$ such that $ABCD$ is a parallelogram. On the basis of the above information, answer the following question.

If $A=(3,5),$ then circumcenter of $△BCD$ is

• A. $(\frac{3}{17},\frac{5}{34})$
• B. $(\frac{5}{17},\frac{3}{34})$
• C. $(\frac{3}{34},\frac{5}{34})$
• D. $(\frac{3}{17},\frac{5}{17})$

Answer 1

The correct option is C $(\frac{3}{34},\frac{5}{34})$

Length of tangents from a point (outside the circle) are equal in length

$\therefore AB=AD$ then the parallelogram $ABCD$ is a rhombus.

$\therefore$ Mid point of $BD=$ mid point of $AC$

$\therefore$ $C$ is mirror image of point $A$ w.r to $BD$, Also $BD\perp AC$
$\because A=(3,5)$

$\therefore$ Equation of $BC$ is rhe equation of chord of contact of tangents drawn from $(3,5)$ to the circle $x^2+y^2=9$ is given by $3x+5y=9$

Let $C(\alpha ,\beta )$ is the image of point $A(3,5)$ the $3x+5y=9$

$\therefore \frac{\alpha -3}{3}=\frac{\beta -5}{5}=-2\frac{(9+25-9)}{34}$

$\therefore \frac{\alpha -3}{3}=\frac{\beta -5}{5}=-\frac{25}{17}$

$\Rightarrow \alpha =-\frac{75}{17}+3,\beta =-\frac{125}{17}+5$

$\therefore \alpha =-\frac{24}{17},\beta =-\frac{40}{17}$

Now equation of circumcircle of $△BCD$ is

$(x^2+y^2-9)+\lambda (3x+5y-9)=0$ which passes through the point $(\alpha ,\beta )$ then

$(\frac{576}{289}+\frac{1600}{289}-9)+\lambda (\frac{-3\times 24}{17}+\frac{-5\times 40}{17}-9)=0$

$\Rightarrow \frac{2176-2601}{289}-\lambda \frac{(72+200+153)}{17}=0$

$\Rightarrow \frac{-425}{289}=\frac{\lambda \left(425\right)}{17}$
$\therefore \lambda =-\frac{1}{17}$

Equation circumcircle of $△BCD$ is

$(x^2+y^2-9)+2x\left(\frac{3\lambda }{2}\right)+2y\left(\frac{5\lambda }{2}\right)-9\lambda =0$

$(x^2+y^2-9)+\left(\frac{-3}{17}\right)x+\left(\frac{-5}{17}\right)y+\frac{9}{17}=0$

$\Rightarrow$ Circumcenter $(\frac{-3}{2}\lambda ,\frac{-5}{2}\lambda )=(\frac{3}{34},\frac{5}{34})$

Hence choice ( c) is correct answer.