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 abscissa of all such points P is __________

Let $$(P(x_1,y_1)$$) lie on the circle and ($$Q(x_2,y_2))$$ on the line.

Circle:
$$x_1^2+y_1^2=4$$

Line for (Q):
$$5x_2+y_2+2=0;\Rightarrow;y_2=-5x_2-2$$

Given perpendicular bisector has slope (1) (since ($$x-y+1=0\Rightarrow y=x+1$$))
So slope of (PQ = -1):
$$\frac{y_2-y_1}{x_2-x_1}=-1\Rightarrow x_2+y_2=x_1+y_1$$

Midpoint lies on bisector

$$\frac{x_1+x_2}{2}-\frac{y_1+y_2}{2}+1=0$$
$$\Rightarrow x_1-y_1+x_2-y_2+2=0$$

From line:
$$x_2+y_2=-4x_2-2$$
$$\Rightarrow-4x_2-2=x_1+y_1$$

Also:
$$x_2-y_2=6x_2+2$$
]Substitute into midpoint equation → simplifies to:

$$x_1=2-5y_1$$

Put into circle:
$$(2-5y)^2+y^2=4$$
$$\Rightarrow y(13y-10)=0$$

$$(y=0\Rightarrow x=2)$$

$$(y=\frac{10}{13}\Rightarrow x=-\frac{24}{13})$$

Sum of abcissas

$$2+\left(-\frac{24}{13}\right)=\frac{2}{13}$$
$$13\times\frac{2}{13}=2$$