A variable circle passes through the fixed `A (p, q)` and touches the x-axis. Show that the locus of the other end of the diameter through `A` is (x-p)^2 = 4qy`.

Options

A. `(y-q)^(2)=4px`
B. `(x-q)^(2)=4py`
C. `(y-p)^(2)=4qx`
D. `(x-p)^(2)=4qy`

Answer

D

Explanation

Let (h, k) be the coordinates of the other end B of the diameter through A. The coordinates of the centre are `((p+h)/(2), (q+k)/(2))`.
Since the circle touches x-axis. Therefore,
|y-coordinates of its centre|=Radius
`rArr |(q+k)/(2)=(1)/(2)sqrt((p-h)^(2)+(q-k)^(2))`
`rArr (q+k)^(2)=(p-h)^(2)+(q-k)^(2)`
`rArr 4qk=(h-p)^(2)`
Hence, the locus of (h, k) is `(x-p)^(2)=4qy`.