Question

A straight line is drawn through the centre of the circle `x^2 + y^2-2ax=0` line `x + 2y = 0 ` parallel to the straight and intersecting the circle at `A` and `B`. Then the area of triangle `AOB` is

Answer 1

`x^2+y^2-2ax=0`
`x+2y=0`
equation of line
x+2y++c=0
a+0+c=0
c=-a
We get,
x+2y-a=0
From diagram
`h=|(-a/sqrt5)`
`h=a/sqrt5`
b=2a
area of triangle=`1/2*B*H`
=`1/2*a/sqrt5*2a`
=`a^2/sqrt5`