For all real values of k,the point which lies | vedclass

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

Question

(A) 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 $T=0$,which is $xx_1+yy_1+g(x+x_1)+

Vedclass · Accepted Answer

$(3, -1)$

Vedclass · Answer

(A) 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 $T=0$,which is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. Given the circle $x^2+y^2+4x-8y-5=0$,we have $g=2, f=-4, c=-5$. The polar of $(k, k+1)$ is: $kx + (k+1)y + 2(x+k) - 4(y+k+1) - 5 = 0$ $kx + ky + y + 2x + 2k - 4y - 4k - 4 - 5 = 0$ $(k+2)x + (k-3)y - 2k - 9 = 0$ Rearranging the terms to isolate $k$: $k(x+y-2) + (2x-3y-9) = 0$ For this to be true for all real values of $k$,both coefficients must be zero: $x+y-2 = 0$ and $2x-3y-9 = 0$. Solving these equations: From the first,$y = 2-x$. Substituting into the second: $2x - 3(2-x) - 9 = 0 \Rightarrow 2x - 6 + 3x - 9 = 0 \Rightarrow 5x = 15 \Rightarrow x = 3$. Then $y = 2-3 = -1$. The point is $(3, -1)$.

For all real values of $k$,the point which lies on the polar of $(k, k+1)$ with respect to the circle $x^2+y^2+4x-8y-5=0$ is

Similar Questions

Let $P$ be any point on the circle $x^2+y^2=25$. Let $L$ be the chord of contact of $P$ with respect to the circle $x^2+y^2=9$. The locus of the poles of the lines $L$ with respect to the circle $x^2+y^2=36$ is

The inverse of the point $(1, 2)$ with respect to the circle $x^2 + y^2 - 4x - 6y + 9 = 0$ is

The pole of the line $\frac{x}{a} + \frac{y}{b} = 1$ with respect to the circle $x^2 + y^2 = c^2$ 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

If $Q$ is the inverse point of the point $P(2, 3)$ with respect to the circle $x^2+y^2-2x-2y+1=0$,then the circle with $PQ$ as diameter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module