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AP EAMCET · Maths · Circle

The point of concurrence of the polars of the variable point \((2 t, t-4), t \in \mathbb{R}\) with respect to the circle \(x^2+y^2-4 x-6 y+1=0\) is

  1. A \((1,3)\)
  2. B \((1,-3)\)
  3. C \((-3,1)\)
  4. D \((3,1)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \((3,1)\)

Step-by-step Solution

Detailed explanation

The polar of \(P(x_1, y_1)\) with respect to \(x^2+y^2+2gx+2fy+c=0\) is \(xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0\). Given \(x_1=2t\), \(y_1=t-4\), \(g=-2\), \(f=-3\), \(c=1\). \(x(2t)+y(t-4)-2(x+2t)-3(y+t-4)+1=0\) \(2tx+ty-4y-2x-4t-3y-3t+12+1=0\) \(t(2x+y-7)+(-2x-7y+13)=0\) Point of…
From AP EAMCET
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