Q91 / 160 · TS EAMCET 2023 (13 May Shift 1) · Mathematics

Let $S \equiv x^2+y^2-8 x+10 y+5=0$ be a circle. Let $P(1,1)$ and $\mathrm{Q}(1,-1)$ be two points. Then the point of intersection of the polar of $\mathrm{P}$ with respect to $\mathrm{S}=0$ and the chord with $\mathrm{Q}$ as mid-point to $\mathrm{S}=0$ is

  • A
    $(2,2)$
  • Correct answer
    B
    $(11,13 / 2)$
  • C
    $(-4,-1)$
  • D
    $(5,7 / 2)$

Solution

$S: x^2+y^2-8 x+10 y+5=0$ $\therefore \quad$ Equation of polar of $P$ w.r.t. $S=0, T=0$ $$ \begin{aligned} & \Rightarrow 1(x)+1(y)-4(x+1)+5(y+1)+5=0 \\ & \Rightarrow-3 x+6 y+6=0 \end{aligned} $$ $\Rightarrow 3 x-6 y-6=0$ ...(1) Equation of chord with $Q$ as mid point $T=S_1$ $$ \begin{aligned} & \Rightarrow x-y-4(x+1)+5(y-1)+5=1+1-8-10+5 \\ & \Rightarrow-3 x+4 y-4=-11 \end{aligned} $$ $\Rightarrow 3 x-4 y-7=0$ ...(2) From (1) and (2), $x=11, y=\frac{13}{2}$

Hence, the correct answer is option: 'B'.

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