If a variable straight line `x cos alpha+y sin alpha=p` which is a chord of hyperbola `(x^(2))/(a^(2))=(y^(2))/(b^(2))=1 (b gt a)` subtends a right angle at the centre of the hyperbola, then it always touches a fixed circle whose radius, is (a) `(sqrt(a^2+b^2))/(ab)` (b) `(2ab)/(sqrt(a^2+b^2))` (c) `(ab)/(sqrt(b^2-a^2))` (d) `(sqrt(a^2+b^2))/(2ab)`

Options

A. `(ab)/(sqrt(b-2a))`
B. `(a)/(a-b)`
C. `(ab)/(sqrt(b^(2)-a^(2)))`
D. `(ab)/(sqrt(b+a))`

Explanation

The combined equation of the straight lines joining, the centre of the hyperbola of the points of intersection of the line `x cosalpha+ysinalpha=p` and the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1` is
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=((xcosalpha+ysinalpha)/(p))^(2)`
This is equation will represent a pair of perpendicular straight lines, if
`(1)/(a^(2))-(cos^(2)alpha)/(p^(2))-(1)/(b^(2))-(sin^(2)alpha)/(p^(2))=0impliesp=(ab)/(sqrt(b^(2)-a^(2)))`
Substituting the value of `p` in `x cosalpha+ysinalpha=p`, we get
`x cosalpha+ysinalpha=(ab)/(sqrt(b^(2)-a^(2)))`
Clearly, it touches the circle `x^(2)+y^(2)=(-(ab)/(sqrt(b^(2))-a^(2))^(2))`