Question

Let $P(3,3)$ and $Q(1,2)$ be two points. If $R$ is a point such that the straight lines $PQ$ and $QR$ are equally inclined to the tangent of the circle $x^2+y^2=5$ at $Q,$ then equation of the line $QR$ is

• A. $2x-11y=15$
• B. $2x+11y=15$
• C. $11x+2y=15$
• D. $11x-2y=15$

Answer 1

The correct option is C $11x+2y=15$

$Q$ lies on the circle $x^2+y^2=5$
Slope of tangent at $Q$ is, $m_t=-\frac{1}{2}$

$PQ$ and $QR$ are making same angle with the tangent line at $Q(1,2)$
and $m_{PQ}=\frac{1}{2}$
Let slope of $QR$ be $m.$
$\left|\frac{\frac{1}{2}+\frac{1}{2}}{1-\frac{1}{4}}\right|=\left|\frac{m+\frac{1}{2}}{1-\frac{m}{2}}\right|$
$\Rightarrow \left|\frac{4}{3}\right|=\left|\frac{2m+1}{2-m}\right|$


$\frac{2m+1}{2-m}=\frac{4}{3}$
$\Rightarrow 6m+3=8-4m$
$\Rightarrow 10m=5$
$\Rightarrow m=\frac{1}{2}$ $($ that is the line $PQ$ itself $)$

$\frac{2m+1}{2-m}=\frac{-4}{3}$
$\Rightarrow 6m+3=-8+4m$
$\Rightarrow 11=-2m$
$\Rightarrow m=-\frac{11}{2}$

Equation of $QR:$
$y-2=-\frac{11}{2}(x-1)$
$\Rightarrow 2y-4=-11x+11$
$\Rightarrow 11x+2y=15$