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x^2+y^2+4x-2y-3=0\\\\(x+2)^2+(y-1)^2=8\\\\Substitute\ x+2=y,\ from\ equation\ of\ line\\\\y^2+(y-1)^2=8\\\\2y^2-2y-7=0\\\\y=\frac{1}{2*2}(2+-\sqrt{4+56})=\frac{1+\sqrt{15}}{2},\ \ or,\ \ \frac{1-\sqrt{15}}{2}\\\\x=\frac{-3+\sqrt{15}}{2},\ \ or,\ \ \frac{-3-\sqrt{15}}{2}\\\\Middle\ point\ is\\\\

( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})\\\\(-\frac{3}{2}, \frac{1}{2})\\

Another way is to find the perpendicular from center of the circle on to the line.

Line perpendicular to x-y+2 =0, is     x+y=c 
It passes through the center of the circle (-2,1)..    

Hence,  -2 + 1 = c   =>    c = -1

Intersection of the two perpendicular  lines y= x+2  and  y = -x - 1  is 
           (-3/2 , 1/2)