Question

If a circle and rectangular hyperbola $xy=c^2$ meet in the four points $t_1,t_2,t_3\&t_4$ . Then which of the following statements are correct?
1. The center of the mean position of the four points bisects the distance between the center of the two curve.
2. Center of the circle through the points $t_1,t_2\&t_3$ is:$\left[\right(\frac{c}{2}(t_1+t_2+t_3)+\frac{1}{t_1.t_2.t_3}),\frac{c}{2}(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+t_1{t}_2{t}_3\left)\right]$

• A. Only 1
• B. Only 2
• C. Both 1 & 2
• D. None of these

Answer 1

The correct option is C

Both 1 & 2

Let the equation of the cicle be

$x^2+y^2+2gx+2fy+c=0$ - - - - - - - (1)

Equation of the rectangular hyperbola is $xy=c^2$

Let insersection point of circle & hyperbola be

$p(c{t}_1,\frac{c}{t_1}),Q(c{t}_2,\frac{c}{t_2}),R(c{t}_3,\frac{c}{t_3}),S(c{t}_4,\frac{c}{t_4})$

solving equations (1) and (2)

substituting $y=\frac{c^2}{x}$ in equation (1)

we get , $x^2+\frac{c^4}{x^2}+2gx+2f\frac{c^2}{x}+c=0$

$x^4+2g{x}^3+c{x}^2+2f{c}^{2}x+c^4=0$ - - - - --- (2)

$\sum x_i=-2g$

similarly when we substitute $x=\frac{c^2}{y}$ in equation (1)

we get , $y^4+2f{y}^3+c{y}^2+2y{c}^{2}y+c^4=0$ - - - - - - -(3)

$\sum yi=-2f$

Mean position of the four points m$(\frac{\sum xi}{4},\frac{\sum yi}{4})$

=$(\frac{-2g}{4},\frac{-2f}{4})$

=$(\frac{-2g}{2},\frac{-f}{2})$

Co-ordinates of centre C is(0,0),centre of the circle (-g,-f)

Midpoint of line joining between centre of the circle R

centre of the rectangular hyperbola is

$(\frac{0-g}{2},\frac{0-f}{2})\equiv (\frac{-g}{2},\frac{-f}{2})$

This is same as the mean position of four points

statement(1) is correct.

statement 2:

centre of the circle is (-g,-f)

From the statement (1)

we can also write it as

$(\frac{\sum xi}{2},\frac{\sum yi}{2})$

$(\frac{x_1+x_2+x_3+x_4}{2},\frac{y_1+y_2+y_3+y_4}{2})$

In the equation (2) we observe that product of the roots

$x_1\times x_2\times x_3\times x_4=c^4$

$c{t}_1.c{t}_2.c{t}_3.c{t}_4=c^4$

$t_1.t_2.t_3.t_4=\frac{c^4}{c^4}=1$

Now,centre of the circle

$(\frac{x_1+x_2+x_3+x_4}{2},\frac{y_1+y_2+y_3+y_4}{2})$

=$(\frac{c{t}_1+c{t}_2+c{t}_3+c{t}_4}{2},\frac{\frac{c}{t_1}+\frac{c}{t_1}+\frac{c}{t_3}+\frac{c}{t_4}}{2})$

=$\left[\frac{c}{2}\right(t_1+t_2+t_3+t_4),\frac{c}{2}(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\frac{1}{t_4}\left)\right]$

$\{using{t}_1{t}_2{t}_3{t}_4=1t_4=\frac{1}{t_1.t_2.t_3}\}$

=$\left[\frac{c}{2}\right(t_1+t_2+t_3+\frac{1}{t_1.t_2.t_3}),\frac{c}{2}(\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+t_1.t_2.t_3\left)\right]$

So,given statement is oorrect.

Both statement (1) and statement (2) are correct.