Find the intervals of the values of `a` for which the line `y+x=0` bisects two chords drawn from the point `((1+sqrt(2)a)/2,(1-sqrt(2)a)/2)` to the circle `2x^2+2y^2-(1+sqrt(2)a)x-(1-sqrt(2)a)=0`

Explanation

Given, `2x^(2) +2y^(2)-(1+sqrt23a)x-(1-sqrt2a)y=0`
`rArr x^(2)+y^(2)-((1+sqrt2a)/(2))x-((1-sqrt2a)/(2))y=0`
Since, y + x = 0 bisects two chords of this circle, mid points of the chords must be of the form `(alpha, - alpha)`.

Equation of the chord having `(alpha, - alpha)` as mid point is
`T=S_(1)`
`rArr xalpha+y(-alpha)-((1+sqrt2a)/(4))(x+alpha)-((1-sqrt2a)/(4))(y-alpha)`
`a^(2)+(-alpha)^(2)-((1+sqrt2a)/(2))alpha-((1-sqrt2a)/(2))(-alpha)`
`rArr 4xalpha-4yalpha-(1+sqrt2alpha)x-(1+sqrt2a)alpha`
`-(1-sqrt2a)y+(1-sqrt2a)alpha`
`4alpha^(2)+4alpha^(2)-(1+sqrt2a)*2alpha+(1-sqrt2a)*2alpha`
`rArr 4ax-45ay-(1+sqrt2a)x-(1-sqrt2a)y`
`=8alpha^(2)-(1+sqrt2a)alpha+(1-sqrt2a)alpha`
But this chord will pass through the point ,brgt `((1+sqrt2a)/(2))f-4alpha((1-sqrt2a)/(2))-((1+sqrt2a)(1+sqrt2a))/(2)`
`-((1-sqrt2a)(1-sqrt2a))/(2)`
`=8alpha^(2)-2sqrt2aalpha`
`rArr 2alpha[(1+sqrt2a-1+sqrt2a)]=8alpha^(2)-2sqrt2aalpha`
`rArr 4sqrt2alpha-(1)/(2)[2+2sqrt2a)^(2)]=8alpha^(2)-2sqrt2aalpha`
`[therefore(a+b)^(2)+(a-b)^(2)=2a^(2)+2b^(2)]`
`rArr 8 alpha^(2)-6sqrt2alpha+1+2a^(2)=0`
But this quadratic equation will have two distinct roots, if
`(6sqrt2a)^(2)-4(8)(1+2a^(2))gt0`
`rArr 72a^(2)-32(1+2a^(2))gt0`
`rArr 8a^(2)-32gt0`
`rArr a^(2)-4gt0`
`rArr a lt-2uuagt2`
Therefore, `a in (-oo, -2)uu(2, oo)`.