A circle of constant radius r passes through the origin O, and cuts the axes at A and B. The locus of the foots the perpendicular from O to AB is `(x^(2) + y^(2))^k =4r^(2)x^(2)y^(2)`, Then the value of k is

Options

A. `(x^(2) + y^(2))(x+y) = R^(2)xy`
B. `(x^(2) + y^(2))^(3) = 4R^(2)x^(2)y^(2)`
C. `(x^(2) + y^(2))^(2) + 4Rx^(2)y^(2)`
D. `(x^(2) + y^(2)) = 4R^(2)x^(2) y^(2)`

Answer

B

Explanation

let A be (a, 0) and B be (0, b) as at a right angle is formed so AB wil act as diameter of circle. So,
`sqrt(a^(2) + b^(2)) = 2R`
`a^(2) + b^(2) = 4R^(2)` …(i)
Let c'(h,k) be required locus.
Also, `angleOC'A = 90^(@)`
So. In `Delta OC' A`
`OA^(2) = (OC')^(2) + (AC')^(2)`
`a^(2) = h^(2) + k^(2) + (h-a)^(2) + (k-0)^(2)`
`a^(2) = h^(2) + k^(2) + h^(2) + a^(2) - 2ah + k^(2)`
`2(h^(2) + k^(2) - ah) = 0`
`a = (h^(2) + k^(2))/(h)` ...(ii)
Similarly `b = (h^(2) + k^(2))/(k)` ...(iii)
Put (ii) & (iii) in (i)
`[(h^(2) + k^(2))/(h)] + [(h^(2)+k^(2))/(k)] = 4R^(2) rArr [h^(2) + k^(2)] = 4R^(2)[h^(2)k^(2)]` Put `h rarr x, k rarr y`