The point of concurrence of all conjugate lines of the line $5 x+7 y-78=0$ with respect to the circle…

The point of concurrence of all conjugate lines of the line $5 x+7 y-78=0$ with respect to the circle $x^2+y^2+6 x+8 y-96=0$ is
  1. (-2 , 3)
  2. (3 ,-2 )
  3. (3 ,2 )
  4. (2, 3)

Solution

The all conjugate lines of the line $5 x+7 y-78=0$ with respect to circle $x^2+y^2+6 x+8 y-96=0$ passes through the pole of the given line with respect to given circle. Let required pole is $P\left(x_1, y_1\right)$, so equation of polar of point $\left(P\left(x_1, y_1\right)\right.$ with respect to given circle is $T=0$. $ \Rightarrow x x_1+y y_1+3\left(x+x_1\right)+4\left(y+y_1\right)-96=0 $
The line (i) represent the line $5 x+7 y-78=0$ only. So, $\frac{x_1+3}{5}=\frac{y_1+4}{7}=\frac{3 x_1+4 y_1-96}{-78}=k$ (Let) $ \Rightarrow x_1=5 k-3, y_1=7 k-4 \text { and } 3 x_1+4 y_1=96-78 k $ So, $\quad 3(5 k-3)+4(7 k-4)=96-78 k$ $ \begin{array}{ll} \Rightarrow & 15 k+28 k+78 k=96+9+16 \\ \Rightarrow & 121 k=121 \Rightarrow k=1 \\ \text { So, } & x_1=2 \text { and } y_1=3 \end{array} $ So, required point of concurrence of all conjugate lines of the given line w.r.t. given circle is $(2,3)$. Hence, option (d) is correct

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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