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Let A (0,2 ) , B ( 3, 0 ) , C ( 6, 4 ) ...

Let `A (0,2 ) , B ( 3, 0 ) , C ( 6, 4 ) ` be the vertices of triangle and P is a point inside the triangle such that area of triangle APB, BPC and CPA are equal. Equation of circle circumscribing ` Delta APB ` is :

A

`x^(2)+y^(2)-2x-3y=0`

B

`x^(2)+y^(2)-4x-3y=0`

C

`x^(2)+y^(2)-3x-4y=0`

D

`x^(2)+y^(2)-3x2y=0`

Text Solution

Verified by Experts

The correct Answer is:
D
`Delta PAB = Delta PBC = DeltaPCA " " rArr " " `P is centroid
P(3,2)
Also `Delta APB` is a night angle .
Circumcirle `rArr (x - 0) (x-3) + (y-2)(y-0)=0`
`rArr " "x^(2) + y^(2) - 3x - 2y = 0`
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