The extremities of a diagonal of a rectangle are `(-4, 4) and (6, -1)`. A circle circumscribes the rectangle and cuts an intercept `AB` on the y-axis. If `Delta` be the area of the triangle formed by `AB` and the tangents to the circle at `A and B`, then `8Delta =`.

Answer

8

Explanation

The circles has the eqation
`(x + 4) (x - 6) + (y - 4)(y + 1) = 0`
i.e. `x^(2) + y^(2) - 2x - 3y - 28 = 0`
`AB` is the intercept of this circle on `y -`axis.
`rArr AB = 2sqrt(f^(2) - c) = 2sqrt((9)/(4) + 28) = 11`
`:'` Both `A, B` lie on `y -` axis
Putting `x = 0` in equation of circle yields.
`y = 7` and `y = -4`
`rArr A = (0, 7)` and `B(0, 4)`
`:.` Tangent at `A: 2x - 11y + 77 = 0`
`:.` Tangent at `B: 2x - 11y + 44 = 0`
They intersect at `C((121)/(4),(3)/(2))`
Hence, area `(DeltaABC) = (1)/(2) xx 11 xx (121)/(4) = (1331)/(8)`
`:. lambda = 1331`