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 $\alpha$ be an integer multiple of $8$. If $S$ is the set of all possible values of $\alpha$ such that the line $6 x + 8 y + \alpha = 0$ intersects the circle $x^2 + y^2 - 4 x - 6 y + 9 = 0$ at two distinct points,then the number of elements in $S$ is

Let the maximum and minimum values of $(\sqrt{8x-x^2-12}-4)^2+(x-7)^2, x \in R$ be $M$ and $m$ respectively. Then $M^2-m^2$ is equal to ...............

If $\theta$ is the angle between the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-8x-12y+43=0$,then $|7 \sec \theta - 18 \cos \theta| = $

In a triangle,two vertices are $(2, 3)$ and $(4, 0)$,and its circumcentre is $(2, z)$ for some real number $z$. The circumradius is

If a circle is inscribed in an equilateral triangle of side $a$,then the area of any square (in sq. units) inscribed in this circle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module