Question

The line $x+y=0$ bisects two chords drawn from the point $(\frac{1+k\sqrt{2}}{2},\frac{1-k\sqrt{2}}{2})$ to the circle $x^2+y^2-\left(\frac{1+k\sqrt{2}}{2}\right)x-\left(\frac{(1-k\sqrt{2})}{2}\right)y=0.$ Then the minimum integral value of $|k|$ is

Answer 1

Let $A(\frac{1+k\sqrt{2}}{2},\frac{1-k\sqrt{2}}{2})\equiv (p,q)$
Equation of chord $AB$ whose mid point is $(h,-h)$ is
$T=S_{1}xh-yh-p\left(\frac{x+h}{2}\right)-q\left(\frac{y-h}{2}\right)=h^2+h^2-ph+qh\Rightarrow 2hx-2hy-p(x+h)-q(y-h)=4h^2-2ph+2qh\Rightarrow 4h^2-ph+qh-x(2h-p)+y(2h+q)=0$
It passes through $A(p,q)$ then
$4h^2+(q-p)h-p(2h-p)+q(2h+q)=0$
Simplifying the above equation
$4h^2-3h(p-q)+p^2+q^2=0$
So for two real and different values of $h$,we must have
$D>0$
$9(p-q{)}^2-16(p^2+q^2)>018k^2-8-16k^2>0k^2-4>0k\in (-\infty ,-2)\cup (2,\infty )$
$\Rightarrow |k|>2$