If 2kx + 3y - 1 = 0 and 2x + y + 5 = 0 are co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the circle is $S \equiv x^2 + y^2 - 2x - 4y - 4 = 0$. The center is $(1, 2)$ and the radius $r = \sqrt{1^2 + 2^2 - (-4)} = \sqrt{9

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The equation of the circle is $S \equiv x^2 + y^2 - 2x - 4y - 4 = 0$. The center is $(1, 2)$ and the radius $r = \sqrt{1^2 + 2^2 - (-4)} = \sqrt{9} = 3$.Two lines $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ are conjugate with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ if the pole of $L_1$ lies on $L_2$.The pole $(x_1, y_1)$ of the line $2kx + 3y - 1 = 0$ with respect to the circle is given by the condition that the chord of contact from $(x_1, y_1)$ is $2kx + 3y - 1 = 0$.The equation of the chord of contact from $(x_1, y_1)$ is $xx_1 + yy_1 - (x + x_1) - 2(y + y_1) - 4 = 0$,which simplifies to $x(x_1 - 1) + y(y_1 - 2) - (x_1 + 2y_1 + 4) = 0$.Comparing this with $2kx + 3y - 1 = 0$,we have $\frac{x_1 - 1}{2k} = \frac{y_1 - 2}{3} = \frac{x_1 + 2y_1 + 4}{1} = \lambda$.So,$x_1 = 2k\lambda + 1$ and $y_1 = 3\lambda + 2$.Since the pole $(x_1, y_1)$ lies on the line $2x + y + 5 = 0$,we substitute these values:$2(2k\lambda + 1) + (3\lambda + 2) + 5 = 0$ $\Rightarrow 4k\lambda + 2 + 3\lambda + 7 = 0$ $\Rightarrow \lambda(4k + 3) = -9$ $\Rightarrow \lambda = \frac{-9}{4k + 3}$.Alternatively,using the condition for conjugate lines $a_1a_2 + b_1b_2 = r^2(l_1l_2 + m_1m_2)$ where lines are $lx+my+n=0$ is not standard. The condition for $L_1, L_2$ being conjugate is $r^2(a_1a_2 + b_1b_2) = (a_1g + b_1f - c_1)(a_2g + b_2f - c_2)$.Here $g = -1, f = -2, c = -4, r^2 = 9$.$9(2k(2) + 3(1)) = (2k(-1) + 3(-2) - (-1))(2(-1) + 1(-2) - 5)$$9(4k + 3) = (-2k - 5)(-9)$$4k + 3 = 2k + 5$ $\Rightarrow 2k = 2$ $\Rightarrow k = 1$.

If $2kx + 3y - 1 = 0$ and $2x + y + 5 = 0$ are conjugate lines with respect to the circle $x^2 + y^2 - 2x - 4y - 4 = 0$,then $k =$

Similar Questions

Consider the circles $S_1: x^2+y^2+2x+8y-23=0$ and $S_2: x^2+y^2-4x+10y+19=0$. If the polars of the centre of one circle with respect to the other circle are $L_1$ and $L_2$,then $L_1$ and $L_2$ are

For all real values of $k$,the polar of the point $(2k, k-4)$ with respect to the circle $x^2+y^2-4x-6y+1=0$ passes through the point:

If the polar of a circle $x^2 + y^2 = a^2$ with respect to a point $(x', y')$ is $Ax + By + C = 0$,then its pole will be:

The polar of $(1, -2)$ with respect to $x^2+y^2-10x-10y+25=0$ is

If the inverse point of $(1,1)$ with respect to the circle $x^2+y^2-4x-6y+12=0$ is $(h, k)$,then $h+k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module