The point of concurrence of all conjugate lin | vedclass

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(D) All conjugate lines of the line $5x + 7y - 78 = 0$ with respect to the circle $x^2 + y^2 + 6x + 8y - 96 = 0$ pass through the pole of the given li

Vedclass · Accepted Answer

$(2, 3)$

Vedclass · Answer

(D) All conjugate lines of the line $5x + 7y - 78 = 0$ with respect to the circle $x^2 + y^2 + 6x + 8y - 96 = 0$ pass through the pole of the given line with respect to the given circle.Let the required pole be $P(x_1, y_1)$. The equation of the polar of point $P(x_1, y_1)$ with respect to the given circle is $T = 0$.$xx_1 + yy_1 + 3(x + x_1) + 4(y + y_1) - 96 = 0$$(x_1 + 3)x + (y_1 + 4)y + (3x_1 + 4y_1 - 96) = 0 \quad \dots (i)$Since line $(i)$ represents the line $5x + 7y - 78 = 0$,we have:$\frac{x_1 + 3}{5} = \frac{y_1 + 4}{7} = \frac{3x_1 + 4y_1 - 96}{-78} = k$ (let)$x_1 = 5k - 3, y_1 = 7k - 4$ and $3x_1 + 4y_1 - 96 = -78k$Substituting $x_1$ and $y_1$ in the third equation:$3(5k - 3) + 4(7k - 4) - 96 = -78k$$15k - 9 + 28k - 16 - 96 = -78k$$43k - 121 = -78k$$121k = 121 \Rightarrow k = 1$Thus,$x_1 = 5(1) - 3 = 2$ and $y_1 = 7(1) - 4 = 3$.The required point of concurrence is $(2, 3)$. Hence,option $(D)$ is correct.

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

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