For all real values of k,the polar of the poi | vedclass

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

Question

(D) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1)

Vedclass · Accepted Answer

$(3, 1)$

Vedclass · Answer

(D) The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2+y^2+2gx+2fy+c=0$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$. Given the circle $x^2+y^2-4x-6y+1=0$,we have $g=-2, f=-3, c=1$. For the point $(2k, k-4)$,the polar equation is: $x(2k) + y(k-4) - 2(x+2k) - 3(y+k-4) + 1 = 0$ $2kx + ky - 4y - 2x - 4k - 3y - 3k + 12 + 1 = 0$ $k(2x + y - 7) - 2x - 7y + 13 = 0$ For this line to pass through a fixed point regardless of $k$,the coefficients of $k$ and the constant term must both be zero: $2x + y - 7 = 0$ $-2x - 7y + 13 = 0$ Adding these equations: $-6y + 6 = 0 \Rightarrow y = 1$. Substituting $y=1$ into $2x + y - 7 = 0$: $2x + 1 - 7 = 0$ $\Rightarrow 2x = 6$ $\Rightarrow x = 3$. Thus,the polar always passes through the point $(3, 1)$.

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:

Similar Questions

If the inverse of $P(-3, 5)$ with respect to a circle is $(1, 3)$,then the polar of $P$ with respect to that circle is

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

The pole of the line $x-5y-7=0$ with respect to the circle $S \equiv x^2+y^2-2x+4y+1=0$ is $P(a, b)$. If $C$ is the centre of the circle $S=0$,then $PC=$

The pole of the straight line $9x + y - 28 = 0$ with respect to the circle $2x^2 + 2y^2 - 3x + 5y - 7 = 0$ is:

The pole of the line $x+y+2=0$ with respect to the circle $x^2+y^2-4x+6y-12=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module