If the pair of lines `ax^2+2(a+b)xy+by^2=0` lie long diameters of a circle and divide the circle into four sectots such that the area of one of the sector is thirce the area of the another sector , then

Options

A. `3a^(2)+2ab+3b^(2)=0`
B. `3a^(2)+10ab+3b^(2)=0`
C. `3a^(2)-2ab+3b^(2)=0`
D. `3a^(2)-10ab+3b^()=0`

Answer

A

Explanation

Given equation is a homogeneous equation of second degree. So, lines represented by it intersect at the origin. Consequently, centre of the circle is at the origin. It is given that the lines represented by the given equation divide the circle into four sectors such that the area of one of the sector in thrice that of the other. Let `theta` be the a cute angle between the lines. Then, `3 theta` is the obtuse angle and so
`theta+3theta=pirArr theta=(pi)/(4)`
`therefore" "tan theta=(2sqrt(h^(2)-ab))/(a+b)`
`rArr" "tan.(pi)/(4)=(2sqrt((a+b)^(2)-ab))/(a+b)`
`rArr" "(a+b)^(2)=4{(a+b)-ab}rArr 3a^(2)+3b^(2)+2ab=0`