The pole of the straight line 9x + y - 28 = 0 | vedclass

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

Question

(B) Let $(h, k)$ be the pole of the line $9x + y - 28 = 0$ with respect to the circle $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$. The

Vedclass · Accepted Answer

$(3, -1)$

Vedclass · Answer

(B) Let $(h, k)$ be the pole of the line $9x + y - 28 = 0$ with respect to the circle $x^2 + y^2 - \frac{3}{2}x + \frac{5}{2}y - \frac{7}{2} = 0$. The equation of the polar of $(h, k)$ with respect to the circle is given by $hx + ky - \frac{3}{4}(x + h) + \frac{5}{4}(y + k) - \frac{7}{2} = 0$. Rearranging the terms,we get $x(h - \frac{3}{4}) + y(k + \frac{5}{4}) - \frac{3}{4}h + \frac{5}{4}k - \frac{7}{2} = 0$,which simplifies to $x(4h - 3) + y(4k + 5) - 3h + 5k - 14 = 0$. Since this equation represents the same line as $9x + y - 28 = 0$,the coefficients must be proportional: $\frac{4h - 3}{9} = \frac{4k + 5}{1} = \frac{-3h + 5k - 14}{-28} = \lambda$. From this,we have $4h - 3 = 9\lambda \Rightarrow h = \frac{9\lambda + 3}{4}$ and $4k + 5 = \lambda \Rightarrow k = \frac{\lambda - 5}{4}$. Substituting these into the third ratio: $-3h + 5k - 14 = -28\lambda$. $-3(\frac{9\lambda + 3}{4}) + 5(\frac{\lambda - 5}{4}) - 14 = -28\lambda$. Multiplying by $4$: $-27\lambda - 9 + 5\lambda - 25 - 56 = -112\lambda$. $-22\lambda - 90 = -112\lambda$ $\Rightarrow 90\lambda = 90$ $\Rightarrow \lambda = 1$. Substituting $\lambda = 1$ back into the expressions for $h$ and $k$: $h = \frac{9(1) + 3}{4} = 3$ and $k = \frac{1 - 5}{4} = -1$. Thus,the pole is $(3, -1)$.

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

Similar Questions

The polar of the point $(5, -1/2)$ with respect to the circle $(x - 2)^2 + y^2 = 4$ is

If $2x - 3y + 3 = 0$ and $x + 2y + k = 0$ are conjugate lines with respect to the circle $S \equiv x^2 + y^2 + 8x - 6y - 24 = 0$,then the length of the tangent drawn from the point $\left(\frac{k}{4}, \frac{k}{3}\right)$ to the circle $S = 0$ is

If the polar of a point $P$ with respect to a circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is $x+2y=4r$,then the point $P$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module