Vertices of a variable triangle are `(3,4); (5costheta, 5sintheta)` and `(5sintheta,-5costheta)` where `theta` is a parameter then the locus of its orthocentre is 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`

Options

A. 1
B. `1//2`
C. 2
D. `3//2`

Answer

A

Explanation

Let the vertices be `O(0,0),A(alpha,0)`, and `B(alpha_1,beta_1)`, where `0le alphale1 ` and `1lealpha_(1)^(2)+beta_(1)^(2) le4`
So, the area of `DeltaOAB` is maximum where `alpha=1` and `(alpha_1,beta_1)` is (2,0)
In this case, `a=1,b=2`, and `c=sqrt5`,which satisfies `2lecle3`. Therefore, the maximum area is 1.