Question

The equation of the chord $AB$ of the circle $x^2+y^2=r^2$ passing through the point $(1,1)$ such that $\frac{PB}{PA}=\frac{\sqrt{2}+r}{\sqrt{2}-r}\cdot (0<r<\sqrt{2})$ is

• A. $x+y=2$
• B. $2x-y=1$
• C. $x+2y=3$
• D. $x-y=0$

Answer 1

Answer: (D)

$x-y=0$

The equation of given circle is,

$x^2+y^2=r^2$

Thus, center of circle is at origin with coordinates $C(0,0)$ and radius is r.

Let chord AB of the circle passes through point $P(1,1)$.

Let $PA=r_1$ and $PB=r_2$

Thus, as per given condition,

$\frac{r_2}{r_1}=\frac{\sqrt{2}+r}{\sqrt{2}-r}$

Now, As per two point form of equation of line, we can write equation of chord AB as,

$y-y_1=m(x-x_1)$

where, $m$ = slope of chord AB.

As chord AB is passing through $P(1,1)$ and $m=tan\theta$, we can write

$y-1=tan\theta (x-1)$

$\therefore y-1=\frac{sin\theta }{cos\theta }(x-1)$

$\therefore cos\theta (y-1)=sin\theta (x-1)$

$\therefore \frac{(x-1)}{cos\theta }=\frac{(y-1)}{sin\theta }$

Let $\frac{(x-1)}{cos\theta }=\frac{(y-1)}{sin\theta }=\acute{r}$

Lets consider equality $\frac{(x-1)}{cos\theta }=\acute{r}$

$\therefore (x-1)=\acute{r}cos\theta$

$\therefore x=\acute{r}cos\theta +1$

Similarly consider equality $(y-1)=\acute{r}sin\theta$

$\therefore y=\acute{r}sin\theta +1$

Now, in parametric form of equation of circle, we can write,

$x^2+y^2=r^2$

$\therefore {(\acute{r}cos\theta +1)}^2+{(\acute{r}sin\theta +1)}^2=r^2$

$\therefore {\acute{r}}^2{cos}^2\theta +2\acute{r}cos\theta +1+{\acute{r}}^2{sin}^2\theta +2\acute{r}sin\theta +1=r^2$

$\therefore {\acute{r}}^2({sin}^2\theta +{cos}^2\theta )+2\acute{r}(cos\theta +sin\theta )+2-r^2=0$

$\therefore {\acute{r}}^2+2\acute{r}(cos\theta +sin\theta )+2-r^2=0$

This is quadratic equation in $\acute{r}$

$a=1$, $b=2(cos\theta +sin\theta )$ and $C=2-r^2$

Let $r_1$ and $r_2$ be two roots of this equation.

$\therefore r_1+r_2=-\frac{b}{a}$

$\therefore r_1+r_2=-\frac{2(cos\theta +sin\theta )}{1}=-2(cos\theta +sin\theta )$ Equation (1)

similarly, $\therefore r_1{r}_2=\frac{c}{a}$

$\therefore r_1{r}_2=\frac{2-r^2}{1}=2-r^2$ Equation (2)

Now, as given in problem,

$\frac{r_2}{r_1}=\frac{\sqrt{2}+r}{\sqrt{2}-r}$

$\therefore r_2=\left(\frac{\sqrt{2}+r}{\sqrt{2}-r}\right)r_1$ Equation (3)

Put this value in equation (1) ,we get,

$r_1+\left(\frac{\sqrt{2}+r}{\sqrt{2}-r}\right)r_1=-2(cos\theta +sin\theta )$

$r_1(1+\frac{\sqrt{2}+r}{\sqrt{2}-r})=-2(cos\theta +sin\theta )$

$\therefore r_1\left(\frac{2\sqrt{2}}{\sqrt{2}-r}\right)=-2(cos\theta +sin\theta )$

$\therefore r_1=\frac{-2(\sqrt{2}-r)(cos\theta +sin\theta )}{2\sqrt{2}}$

$\therefore r_1=\frac{-(\sqrt{2}-r)(cos\theta +sin\theta )}{\sqrt{2}}$

Put this value in equation (2) and equation (3), we get,

$\left[\left(\frac{\sqrt{2}+r}{\sqrt{2}-r}\right)\right]{\left[\frac{-(\sqrt{2}-r)(cos\theta +sin\theta )}{\sqrt{2}}\right]}^2=2-r^2$

$\left[\left(\frac{\sqrt{2}+r}{\sqrt{2}-r}\right)\right]\left(\frac{{(\sqrt{2}-r)}^2{(cos\theta +sin\theta )}^2}{2}\right)=2-r^2$

Simplifying above equation, we get,

$(\sqrt{2}+r)\left(\frac{(\sqrt{2}-r)({cos}^2\theta +si{n}^2\theta +2cos\theta sin\theta )}{2}\right)=2-r^2$

$\therefore \left(\frac{({\left(\sqrt{2}\right)}^2-r^2)(1+sin2\theta )}{2}\right)=2-r^2$

$\therefore (2-r^2)(1+sin2\theta )=2(2-r^2)$

$\therefore (1+sin2\theta )=2$

$\therefore sin2\theta =2-1=1$

$\therefore 2\theta ={sin}^{-1}\left(1\right)$

$\therefore 2\theta ={90}^0$

$\therefore \theta ={45}^0$

Thus, $m=tan\left({45}^0\right)$

$\therefore m=1$

Thus, equation of chord AB passing through $(1,1)$ is given by,

$y-1=1(x-1)$

$\therefore y-1=x-1$

$\therefore y=x$

$\therefore x-y=0$

Thus answer is option (D)