The polars of (-1, 2) with respect to the two | vedclass

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

Question

(D) The polar of the circle $x^2+y^2+2gx+2fy+c=0$ with respect to the pole $(x_1, y_1)$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. For $S_1 \equiv

Vedclass · Accepted Answer

Intersecting at a non-zero point

Vedclass · Answer

(D) The polar of the circle $x^2+y^2+2gx+2fy+c=0$ with respect to the pole $(x_1, y_1)$ is given by $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$. For $S_1 \equiv x^2+y^2+6y+7=0$ with pole $(-1, 2)$,the polar is: $-x+2y+3(y+2)+7=0 \Rightarrow -x+5y+13=0$ (Equation $1$). For $S_2 \equiv x^2+y^2+6x+1=0$ with pole $(-1, 2)$,the polar is: $-x+2y+3(x-1)+1=0$ $\Rightarrow 2x+2y-2=0$ $\Rightarrow x+y-1=0$ (Equation $2$). To check the intersection,we solve the system: $-x+5y+13=0$ $x+y-1=0$ Adding these equations gives $6y+12=0$,so $y=-2$. Substituting $y=-2$ into $x+y-1=0$ gives $x-2-1=0$,so $x=3$. The polars intersect at the point $(3, -2)$,which is a non-zero point.

The polars of $(-1, 2)$ with respect to the two circles $S_1 \equiv x^2+y^2+6y+7=0$ and $S_2 \equiv x^2+y^2+6x+1=0$ are

Similar Questions

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

If the inverse point of the point $(-1, 1)$ with respect to the circle $x^2+y^2-2x+2y-1=0$ is $(p, q)$,then $p^2+q^2=$

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

If $2x - 3y + 1 = 0$ is the equation of the polar of a point $P(x_1, y_1)$ with respect to the circle $x^2 + y^2 - 2x + 4y + 3 = 0$,then $3x_1 - y_1 =$

If the pole of the line $3x - 16y + 48 = 0$ with respect to the hyperbola $9x^2 - 16y^2 = 144$ is $(\alpha, \beta)$,then $\alpha - \beta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module