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
- A \((1,3)\)
- B \((1,-3)\)
- C \((-3,1)\)
- D \((3,1)\)
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…
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