Question

Observe the following For the circle
$S\equiv x^2+y^2=a^2$

List I	List II
$A)$ The pole of the line $x+my+n=0$ is	1. $T=S_1$
$B)$ The inverse point of 2 $(x_1,y_1)$ is	2. $\left(\frac{a^2{x}_1}{x_1^2+y_1^2},\frac{a^2{y}_1}{x_1^2+y_1^2})\right)$
$C)$ The mid point of chord $lx+my+n=0$ is	3. $\left(\frac{-\ln }{l^2+m^2},\frac{-mn}{l^2+m^2}\right)$
$D)$ Equation of the chord of the circle with $(x_1,y_1)$ as the mid-point is	4. $\left(\frac{-a^{2}l}{n},\frac{-a^{2}m}{n}\right)$

• A. $4,1,2,3$
• B. $2,3,1,4$
• C. $4,2,3,1$
• D. $1,3,2,4$

Answer 1

Answer: (C)

$4,2,3,1$

(A) Let $P(h,k)$ be a pole of $ln+my+n=0$
then $xh+ky=a^2$ and $lx+my+n=0$ represent same line
$\therefore \frac{h}{l}=\frac{k}{m}=\frac{-a^2}{n}$
$\therefore h=\frac{-l{a}^2}{n}andk=\frac{-m{a}^2}{n}$
B) Let $p(h,k)$ is inverse point of
$Q(x_1,y_1)$ w.r.t. circle with
centre at $O(0,0)$ and radius as $=OQ\cdot OP=r^2=a^2$
$\sqrt{x_1^2+y_1^2}$ $\cdot \sqrt{h^2+k^2}=a^2$
$h^2+k^2=\frac{a^4}{x_1^2+y_1^2}$
and lies on OQ line
$E{q}^n$ of OQ is $y=\frac{y_1}{x_1}x$
$Q(h,\frac{y_{1}h}{x_1}),then{h}^2\left(\frac{x_1^2+y_1^2}{x_1^2}\right)=\frac{a^4}{x_1^2+y_1^2}$
$h=\frac{x_1{a}^2}{x_1^2+y_1^2}$
and $k=\frac{y_1{a}^2}{x_1^2+y_2^2}$
C) foot of perpendicular on any chord from centre of circle is mid-point of chord.
$\therefore$ let mid-point of $lk+my+h=0$ is $p(h,k)$
$\frac{h-0}{l}=\frac{k-0}{m}=-\left(\frac{h}{l^2+m^2}\right)$
$h=\frac{-hl}{l^2+m^2}$ and $k=\frac{-mn}{l^2+h^2}$
D) $e{q}^n$ of chord with mid point $(x_1,y_1)$ is $T=S_1$