If P is a point on the circle x^{2}+y^{2}=4, | vedclass

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

Question

(B) Let $P = (2 \cos 	heta, 2 \sin 	heta)$ and $Q = (\alpha, -5\alpha - 2)$.Since $x-y+1=0$ is the perpendicular bisector of $PQ$, the midpoint of $

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $P = (2 \cos 	heta, 2 \sin 	heta)$ and $Q = (\alpha, -5\alpha - 2)$.Since $x-y+1=0$ is the perpendicular bisector of $PQ$, the midpoint of $PQ$ lies on the line $x-y+1=0$.Midpoint $M = (\frac{2 \cos 	heta + \alpha}{2}, \frac{2 \sin 	heta - 5\alpha - 2}{2})$.Substituting $M$ into $x-y+1=0$:$\frac{2 \cos 	heta + \alpha}{2} - \frac{2 \sin 	heta - 5\alpha - 2}{2} + 1 = 0$$2 \cos 	heta + \alpha - 2 \sin 	heta + 5\alpha + 2 + 2 = 0$$\cos 	heta - \sin 	heta + 3\alpha + 2 = 0 \quad \dots(1)$Also, the slope of $PQ$ must be the negative reciprocal of the slope of $x-y+1=0$ (which is $1$). Thus, the slope of $PQ$ is $-1$.$\frac{(2 \sin 	heta) - (-5\alpha - 2)}{2 \cos 	heta - \alpha} = -1$$2 \sin 	heta + 5\alpha + 2 = -2 \cos 	heta + \alpha$$\sin 	heta + \cos 	heta + 2\alpha + 1 = 0 \quad \dots(2)$From $(1)$, $3\alpha = \sin 	heta - \cos 	heta - 2$. From $(2)$, $2\alpha = -\sin 	heta - \cos 	heta - 1$.Equating $\alpha$: $\frac{\sin 	heta - \cos 	heta - 2}{3} = \frac{-\sin 	heta - \cos 	heta - 1}{2}$$2 \sin 	heta - 2 \cos 	heta - 4 = -3 \sin 	heta - 3 \cos 	heta - 3$$5 \sin 	heta + \cos 	heta = 1$Using $R \cos(	heta - \phi) = 1$ where $R = \sqrt{26}$, or solving for $\cos 	heta$:$5 \sin 	heta = 1 - \cos 	heta \implies 25(1 - \cos^2 	heta) = (1 - \cos 	heta)^2$$25(1 - \cos 	heta)(1 + \cos 	heta) = (1 - \cos 	heta)^2$Case $1$: $1 - \cos 	heta = 0 \implies \cos 	heta = 1$. Then $P_x = 2(1) = 2$.Case $2$: $25(1 + \cos 	heta) = 1 - \cos 	heta \implies 26 \cos 	heta = -24 \implies \cos 	heta = -\frac{12}{13}$. Then $P_x = 2(-\frac{12}{13}) = -\frac{24}{13}$.Sum of abscissae $= 2 - \frac{24}{13} = \frac{26-24}{13} = \frac{2}{13}$.$13 	imes (	ext{Sum}) = 13 	imes \frac{2}{13} = 2$.

If $P$ is a point on the circle $x^{2}+y^{2}=4$, $Q$ is a point on the straight line $5x+y+2=0$ and $x-y+1=0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of the abscissae of all such points $P$ is ........... .

Similar Questions

Let $C$ be the circle with centre at $(1, 1)$ and radius $= 1$. If $T$ is the circle centred at $(0, y)$,passing through the origin and touching the circle $C$ externally,then the radius of $T$ is equal to:

The line $y=mx+c$ intersects the circle $x^2+y^2=r^2$ at two distinct points if:

$A$ square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is

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

The curve $xy = c, (c > 0)$,and the circle $x^2 + y^2 = 1$ touch at two points. Then the distance between the points of contact is

Vedclass Test Series

Exam Paper Generator

Online Exam Module