If P is a point on the circle x^2+y^2=4, Q is a point on the straight line 5 x+y+2=0 and x-y+1=0 is the perpendicular bisector of PQ, then 13 times…

Mid point of PQ lies on $x - y + 1 = 0$x−y+1=0

$$\dfrac{2\cos\theta + \alpha}{2} - \dfrac{2\sin\theta - 5\alpha - 2}{2} + 1 = 0$$22cosθ+α​−22sinθ−5α−2​+1=0

$$2\cos\theta + \alpha - 2\sin\theta + 5\alpha + 2 + 2 = 0$$2cosθ+α−2sinθ+5α+2+2=0

$\cos\theta - \sin\theta + 3\alpha + 2 = 0$cosθ−sinθ+3α+2=0 ...(1)

$\because$∵ Slope of PQ is $-1$−1

$$\dfrac{2\sin\theta + 5\alpha + 2}{2\cos\theta - \alpha} = -1$$2cosθ−α2sinθ+5α+2​=−1

$$2\sin\theta + 5\alpha + 2 = -2\cos\theta + \alpha$$2sinθ+5α+2=−2cosθ+α

$\sin\theta + \cos\theta + 2\alpha + 1 = 0$sinθ+cosθ+2α+1=0 ...(2)

eliminate $\alpha$α from (1) and (2)

$\Rightarrow \cos\theta + 5\sin\theta = 1$⇒cosθ+5sinθ=1 , $\theta \in [0, 2\pi]$θ∈[0,2π]

$$\Rightarrow 5 \times 2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2} = 2\sin^2\dfrac{\theta}{2}$$⇒5×2sin2θ​cos2θ​=2sin22θ​

$$\therefore \sin\dfrac{\theta}{2} = 0 \Rightarrow \cos\theta = 1$$∴sin2θ​=0⇒cosθ=1

or

$$\sin\dfrac{\theta}{2} = 5 \Rightarrow \cos\theta = -\dfrac{12}{13}$$sin2θ​=5⇒cosθ=−1312​

Sum of all possible values of abscissa of point P is

$$= 2 \times 1 + 2\left(\dfrac{-12}{13}\right) = \dfrac{2}{13}$$=2×1+2(13−12​)=132​

$\therefore$∴ 13 times sum of all possible values of abscissa of point P is 2.