If a variable line has its intercepts on the coordinate axes `e and e^(prime),` where `e/2a n d e^(prime)/2` are the eccentricities of a hyperbola and its conjugate hyperbola, then the line always touches the circle `x^2+y^2=r^2,` where `r=`

Options

A. 1
B. 2
C. 3
D. cannot be decided

Answer

C

Explanation

Since `e//2` and `e'//2` are the eccentricities of a hyperbola and its conjuate hyperbola, we have
`(4)/(e^(2))+(4)/(e'^(4))=1`
`"or "4=(e^(2)e'^(2))/(e'^(2)+e'^(2))`
The line passing through the points (e,0) and (0, e') is
`e'x+ey-ee'=0`
It is tangent to the circle `x^(2)+y^(2)=r^(2)`. Hence,
`(ee')/(sqrt(e^(2)+e'^(2)))=r`
`"or "r^(2)=(e^(2)e'^(2))/(e^(2)+e'^(2))=4`
`"or "r=2`