Chord $x+y=1$ Subtends $45^{\circ}$ at the centre If the Chord $x+y=1$ Subtends $45^{\circ}$ at the centre of the circle $x^2+y^2=a^2$, Find $a$
Method $1$:
We have radius=$a$
If $O$ is the centre and chord be $AB$ we have
$$\angle AOB =45$$
If $OM$ is drawn perpendicular to the chord we have:
$$\angle AOM=22.5$$
In $\Delta AOM$ we have
$$cos(22.5)=\frac{OM}{a}$$
$OM$ is perpendicular distance from $(0,0)$ to $x+y=1$ which is
$$OM=\left|\frac{0+0-1}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$$
Hence we get $$a=\sqrt{2-\sqrt{2}}$$
Method $2.$
By Homogenization of line and curve we have:
$$x^+y^2=a^2(x+y)^2$$
$$(a^2-1)x^2+2xya^2+(a^2-1)y^2=0$$
This pair of straight lines subtends angle of $45^{\circ}$
We have formula that:
$$\tan (\theta)=\left|\frac{2\sqrt{H^2-AB}}{A+B}\right|$$
We have
$A=a^2-1=B$ and $H=a^2$
We get
$$1=\left|\frac{2\sqrt{a^4-(a^2-1)^2}}{2(a^2-1)}\right|$$
Simplifying we get
$$a^4-4a^2+2=0$$
Giving
$$a=\sqrt{2\pm \sqrt{2}}$$
Why in first Method, i could not get second answer?

It's not that you couldn't get the second answer of the second method $\sqrt{2+\sqrt2}$ in the first method. It's that $\sqrt{2+\sqrt2}$ is wrong – drawing the associated circle and line and measuring the subtended angle gives you well over $90^\circ$. There is only one (positive) solution $a$.
