If the circle `x^2+y^2=a^2` intersects the hyperbola `x y=c^2` at four points `P(x_1, y_1),Q(x_2, y_2),R(x_3, y_3),` and `S(x_4, y_4),` then `x_1+x_2+x_3+x_4=0` `y_1+y_2+y_3+y_4=0` `x_1x_2x_3x_4=C^4` `y_1y_2y_3y_4=C^4`

Options

A. `x_(1)+x_(2)+x_(3)+x_(4)=0`
B. `y_(1)+y_(2)+y_(3)+y_(4)=0`
C. `x_(1)x_(2)x_(3)x_(4)=c^(4)`
D. `y_(1)y_(2)y_(3)y_(4)=c^(4)`

Answer

1,2,3,4

Explanation

Putting `y=c^(2)//x` in `x^(2)+y^(2)=a^(2)`, we get
`x^(2)+(c^(4))/(x^(2))=a^(2)`
or `x^(4)-a^(2)x^(2)+c^(4)=0`
As `x_(1),x_(2),x_(3)` and `x_(4)` are the roots of (i) ,we have
`x_(1)+x_(2)+x_(3)+x_(4)=0` and `x_(1)x_(2)x_(3)x_(4)=c^(4)`
Similarly, forming equation in , we get
`y_(1)+y_(2)+y_(3)+y_(4)=0` and `y_(1)y_(2)y_(3)y_(4)=c^(4)`