Question

Find the intervals of values of 'a' for which the line y+x=0 bisects two chords drawn from a point ,
$( \frac{1 + \sqrt{2} a}{2} , \: \frac{1 - \sqrt{2}a }{2} )$
to the circle
$2 {x}^{2} + 2 {y}^{2} - (1 + \sqrt{2} a)x - (1 - \sqrt{2} a)y = 0$

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Answer 1

The equation of the circle is,

$\longrightarrow\sf{2x^2+2y^2-(1+a\sqrt2)x-(1-a\sqrt2)y=0}$

Dividing by 2,

$\longrightarrow\sf{x^2+y^2-\left(\dfrac{1+a\sqrt2}{2}\right)x-\left(\dfrac{1-a\sqrt2}{2}\right)y=0}$

Adding $\sf{\dfrac{1+2a^2}{8}}$ to both sides,

$\longrightarrow\sf{x^2+y^2-\left(\dfrac{1+a\sqrt2}{2}\right)x-\left(\dfrac{1-a\sqrt2}{2}\right)y+\dfrac{1+2a^2}{8}=\dfrac{1+2a^2}{8}}$

$\longrightarrow\sf{\left(x-\dfrac{1+a\sqrt2}{4}\right)^2+\left(y-\dfrac{1-a\sqrt2}{4}\right)^2=\dfrac{2a^2+1}{8}}$

Now the equation is in the form $\sf{(x-\alpha)^2+(y-\beta)^2=r^2,}$ where $(\alpha,\ \beta)$ is the center of the circle of radius $\sf{r.}$

Thus we get,

$\longrightarrow\sf{(\alpha,\ \beta)=\left(\dfrac{1+a\sqrt2}{4},\ \dfrac{1-a\sqrt2}{4}\right)}$

$\longrightarrow\sf{r=\dfrac{\sqrt{2a^2+1}}{2\sqrt2}}$

We can see that one solution to the equation of the circle is $\sf{\left(\dfrac{1+a\sqrt2}{2},\ \dfrac{1-a\sqrt2}{2}\right).}$ Hence it is a point on the circle, from which two chords are drawn to the circle which are bisected by $\sf{x+y=0,}$ whose points are in the form $\sf{(k,\ -k).}$

In the fig., BD and BE are such two chords where $\sf{B\left(\dfrac{1+a\sqrt2}{2},\ \dfrac{1-a\sqrt2}{2}\right),}$ and $\sf{A\left(\dfrac{1+a\sqrt2}{4},\ \dfrac{1-a\sqrt2}{4}\right)}$ is the center of the circle. $\sf{F(k,\ -k)}$ is the midpoint of the chord BE which lies on the line $\sf{x+y=0.}$

Since $\sf{AF\perp BE,}$ their slopes give a product of -1.

$\longrightarrow\sf{\left(\dfrac{\dfrac{1-a\sqrt2}{4}+k}{\dfrac{1+a\sqrt2}{4}-k}\right)\left(\dfrac{\dfrac{1-a\sqrt2}{2}+k}{\dfrac{1+a\sqrt2}{2}-k}\right)=-1}$

$\longrightarrow\sf{\left(\dfrac{1+4k-a\sqrt2}{1-4k+a\sqrt2}\right)\left(\dfrac{1+2k-a\sqrt2}{1-2k+a\sqrt2}\right)=-1}$

$\longrightarrow\sf{\dfrac{1+4k-a\sqrt2}{1-4k+a\sqrt2}=\dfrac{2k-a\sqrt2-1}{2k-a\sqrt2+1}}$

$\longrightarrow\sf{\dfrac{1+(4k-a\sqrt2)}{1-(4k-a\sqrt2)}=\dfrac{(2k-a\sqrt2)-1}{(2k-a\sqrt2)+1}}$

So by rule of componendo and dividendo,

$\longrightarrow\sf{4k-a\sqrt2=-\dfrac{1}{2k-a\sqrt2}}$

$\longrightarrow\sf{(4k-a\sqrt2)(2k-a\sqrt2)=-1}$

$\longrightarrow\sf{8k^2-6ak\sqrt2+(2a^2+1)=0}$

This is a quadratic equation in two variables $\sf{a}$ and $\sf{k,}$ whose discriminant should be non - negative. Thus with respect to $\sf{k,}$

$\longrightarrow\sf{(-6a\sqrt2)^2-(4\times8(2a^2+1))\geq0}$

$\longrightarrow\sf{72a^2-32(2a^2+1)\geq0}$

$\longrightarrow\sf{8a^2-32\geq0}$

$\longrightarrow\sf{a^2-4\geq0}$

$\longrightarrow\sf{(a-2)(a+2)\geq0}$

$\Longrightarrow\sf{a\leq-2\quad OR\quad a\geq2}$

Therefore,

$\longrightarrow\large\textsf{ \sf{\underline{\underline{a\in(-\infty,\ -2\,]\cup[\,2,\ \infty)}}} }$

Answer 2

hope thsi will help uu..//##