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TS EAMCET · Maths · Circle

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

  1. A \((2,2)\)
  2. B \((11,13 / 2)\)
  3. C \((-4,-1)\)
  4. D \((5,7 / 2)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \((11,13 / 2)\)

Step-by-step Solution

Detailed explanation

\(S: x^2+y^2-8 x+10 y+5=0\) \(\therefore \quad\) 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…