Vertices of a variable triangle are `(3, 4), (5 cos theta, 5 sin theta)` and `(5 sin theta, -5 cos theta)`, where `theta in R`. Locus of its orthocentre is

Options

A. `(x + y -1)^(2) + (x- y - 7)^(2) = 100`
B. `( x + y -7)^(2) + ( x- y -1)^(2) = 100`
C. `(x + y -7)^(2) + (x + y -1)^(2) = 100`
D. `(x + y -7)^(2) + (x - y + 1)^(2) = 100`

Explanation

We observe that OA = OB = OC . Therefore , O(0,0) is the circumcentre of `Delta` ABC . Let O' (h,k) be the coordinates of its orthocentre.
The centroid G `((3 + 5 cos theta + 5 sin theta)/(3) , (4 + 5 sin theta - 5 cos theta)/(3))` divides the line segment joining OO' in the ratio `1 : 2`
`therefore (1 xx h + 2 xx 0)/(1 + 2) = (3 + 5 cos theta + 5 cos theta)/(3)`
and ,
`(1 xx k + 2 xx 0)/(1 + 2) = ( 4 + 5 sin theta - 5 cos theta)/(3)`

`implies h = 3 + 5 cos theta + 5 sin theta ` and k = 4 + 5 `sin theta - 5 cos theta`
`implies h + k - 7 = 10 sin theta and h - k + 1 = 10 cos theta `
`implies ( h + k - 7)^(2) - (h - k + 1)^(2) = 100`
Hence , the locus of (h,k) is `(x + y -7)^(2) + (x - y + 1)^(2) = 100`.