The locus of the foot of the normal drawn from any point `P(alpha, beta)` to the family of circles `x^(2)+y^(2)-2gx+c=0`, where g is a parameter, is

Options

A. `(x^(2)+y^(2)+c)(y-beta)=2(ya-xbeta) x`
B. `(x^(2)+y^(2)+c)(x-beta)=2(y alpha-x beta)x`
C. `(x^(2)+y^(2)+c)(x-beta)=2(xalpha-y beta)x`
D. none of these

Answer

A

Explanation

Let A (h, k) be the foot of the normal drawn from `P(alpha, beta)` to the given circle. Then, P, Q and C are collinear points.

`|(alpha ,beta,1),(h,k,1),(g,0,1)|=0`
`rArr alphak+beta g - beta h-gk=0 rArr g (beta-k)=beta h - alpha k " " ...(i)`
Also, CQ=Radius
`rArr (g-h)^(2)+k^(2)=g^(2)-c`
`rArr -2gh+h^(2)+k^(2)=-c rArr g=(h^(2)+k^(2)+c)/(2h) " " ...(ii)`
From (i) and (ii), we get `(h^(2)+k^(2)+c)/(2h)=(beta h - alpha k)/(beta-k)`
Hence, the locus of (h, k) is
` (x^(2)+y^(2)+c)(beta-y)=2(beta x-alpha y)x`
Thus, option (a) is correct.