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Step-by-Step Solution
Step 1: Identify the Angle Between the Lines
The given pair of lines is represented by the homogeneous equation
$ a x^2 + 2(a + b)\,x\,y + b y^2 = 0 $.
In the standard form $ A x^2 + 2H x y + B y^2 = 0 $, we have:
$A = a$
$H = a + b$
$B = b$
The angle $ \phi $ between the two lines is given by
$ \tan \phi = \frac{2 \sqrt{H^2 - A B}}{A + B}. $
Step 2: Substitute the Given Coefficients
Substitute $A = a$, $B = b$, and $H = a + b$:
$ H^2 - A B = (a + b)^2 - a b = a^2 + 2ab + b^2 - ab = a^2 + ab + b^2. $
So,
$ \tan \phi = \frac{2 \sqrt{a^2 + ab + b^2}}{a + b}. $
Step 3: Use the Given Condition on the Sector Areas
Because the lines are diameters of a circle centered at the origin, they divide the circle into four sectors. We are told that one sector's area is thrice that of another sector. If $ \phi $ is the angle between the lines (measured at the center), the four sectors have angles
$ \phi,\, \pi - \phi,\, \phi,\, \pi - \phi $.
For one sector to be three times another, we must have either:
$ \phi = 3(\pi - \phi)
\quad \text{or} \quad
\pi - \phi = 3\phi. $
Solving either of these gives
$ \phi = \frac{\pi}{4} \text{ or } \phi = \frac{3\pi}{4}. $
Thus
$ \tan \phi = \pm 1. $
Step 4: Impose the Condition $ \tan \phi = \pm 1 $
From
$ \tan \phi = \frac{2 \sqrt{a^2 + ab + b^2}}{a + b} = \pm 1, $
we get
$ 2 \sqrt{a^2 + ab + b^2} = \pm (a + b). $
Squaring both sides yields:
$ 4 (a^2 + ab + b^2) = (a + b)^2. $
Expand and simplify:
$ 4(a^2 + ab + b^2) = a^2 + 2ab + b^2. $
$ 4a^2 + 4ab + 4b^2 = a^2 + 2ab + b^2. $
$ 3a^2 + 2ab + 3b^2 = 0. $
Step 5: Conclude the Correct Relation
Hence, the required relation is
$ 3a^2 + 2ab + 3b^2 = 0, $
which matches the correct answer given in the options.
