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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 |
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Answer» `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` |
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