Key Concepts and Formulas

• Angle between a pair of lines: The angle $\theta$ between the pair of lines represented by the homogeneous equation $Ax^2 + 2Hxy + By^2 = 0$ is given by: $\tan \theta = \frac{2\sqrt{H^2 - AB}}{|A+B|}$
• Sector Area of a Circle: The area of a sector of a circle with radius $r$ and central angle $\theta$ (in radians) is given by: $Area = \frac{1}{2}r^2\theta$
• Relationship between angles: If a circle is divided into sectors, the sum of the central angles of all sectors is $2\pi$ radians (or 360 degrees).

Step-by-Step Solution

Step 1: Define the angles and set up the area relationship.

Let $\theta$ be the angle of one sector and $3\theta$ be the angle of the other sector. Since the two lines are diameters, they divide the circle into four sectors. Two sectors have angle $\theta$ and the other two have angle $3\theta$. The sum of all angles must be $2\pi$. Therefore, $2\theta + 2(3\theta) = 2\pi$

Simplifying, we get: $2\theta + 6\theta = 2\pi$ $8\theta = 2\pi$ $\theta = \frac{\pi}{4}$

Thus, the two angles between the lines are $\theta = \frac{\pi}{4}$ and $3\theta = \frac{3\pi}{4}$. Since we are looking for the acute angle between the lines, we choose $\frac{\pi}{4}$.

Step 2: Apply the formula for the angle between the pair of lines.

The given equation is $ax^2 + 2(a+b)xy + by^2 = 0$. Comparing this with the general form $Ax^2 + 2Hxy + By^2 = 0$, we have $A = a$, $H = a+b$, and $B = b$. Therefore, the angle $\phi$ between the lines is given by:

$\tan \phi = \frac{2\sqrt{(a+b)^2 - ab}}{|a+b|}$

Since $\phi = \frac{\pi}{4}$, we have $\tan \frac{\pi}{4} = 1$. Substituting this into the equation above:

$1 = \frac{2\sqrt{(a+b)^2 - ab}}{|a+b|}$ $1 = \frac{2\sqrt{a^2 + 2ab + b^2 - ab}}{|a+b|}$ $1 = \frac{2\sqrt{a^2 + ab + b^2}}{|a+b|}$

Step 3: Square both sides and simplify.

Squaring both sides of the equation, we get:

$1 = \frac{4(a^2 + ab + b^2)}{(a+b)^2}$ $(a+b)^2 = 4(a^2 + ab + b^2)$ $a^2 + 2ab + b^2 = 4a^2 + 4ab + 4b^2$ $0 = 3a^2 + 2ab + 3b^2$

Step 4: Consider the other possible angle.

If we consider the obtuse angle between the lines to be $\frac{3\pi}{4}$, then $\tan(\frac{3\pi}{4}) = -1$. This would give us:

$-1 = \frac{2\sqrt{a^2 + ab + b^2}}{a+b}$ $-(a+b) = 2\sqrt{a^2 + ab + b^2}$

Squaring both sides gives: $(a+b)^2 = 4(a^2 + ab + b^2)$ $a^2 + 2ab + b^2 = 4a^2 + 4ab + 4b^2$ $0 = 3a^2 + 2ab + 3b^2$

Step 5: Examine the given options.

Now, we need to consider the case where the absolute value in the denominator $|a+b|$ is handled differently. We have two cases to consider:

Case 1: $a+b > 0$, then we have $3a^2 + 2ab + 3b^2 = 0$ Case 2: $a+b < 0$, then we have $-(a+b) = 2\sqrt{a^2+ab+b^2}$ Squaring both sides, we get $3a^2 + 2ab + 3b^2 = 0$

However, none of the options match this result. Let's revisit step 2 and consider the reciprocal of the tangent. If the angle is $\pi/4$, then $\cot(\pi/4) = 1$. If the angle is $3\pi/4$, then $\cot(3\pi/4) = -1$. So we also have $\cot \phi = \frac{|A+B|}{2\sqrt{H^2 - AB}} = \frac{|a+b|}{2\sqrt{(a+b)^2 - ab}}$ $1 = \frac{|a+b|}{2\sqrt{a^2+ab+b^2}}$ Squaring both sides, $4(a^2+ab+b^2) = (a+b)^2 = a^2+2ab+b^2$ $3a^2 + 2ab + 3b^2 = 0$ Still not the answer. We are given that one sector is thrice the area of the other. This means that the two possible angles formed are $\theta$ and $3\theta$. They must be supplementary. i.e., $\theta + 3\theta = \pi$ or $\theta = \pi/4$. The angle between them is $\pi/4$. Let's assume the angle between the lines is $\alpha$. Then $\tan(\alpha) = \frac{2\sqrt{(a+b)^2-ab}}{|a+b|}$. We also have $\tan(\pi/4) = 1$, so $\frac{2\sqrt{(a+b)^2-ab}}{|a+b|} = 1$. Instead, let us assume that the angle between the two diameters is $\theta$. Then the areas of the sectors are proportional to $\theta$ and $\pi - \theta$. We have $\pi - \theta = 3\theta$, so $4\theta = \pi$ and $\theta = \pi/4$. Or $\theta = 3(\pi - \theta)$, so $\theta = 3\pi - 3\theta$, so $4\theta = 3\pi$, and $\theta = 3\pi/4$.

Thus, the angle between the lines is $\pi/4$. We have $\tan \theta = \frac{2\sqrt{H^2 - AB}}{A+B}$ where $A=a, B=b, H=a+b$. Then $\tan \theta = \frac{2\sqrt{(a+b)^2 - ab}}{a+b} = \frac{2\sqrt{a^2+ab+b^2}}{a+b}$. If $\theta = \pi/4$, then $\tan \theta = 1$, so $4(a^2+ab+b^2) = (a+b)^2 = a^2+2ab+b^2$. Thus $3a^2+2ab+3b^2=0$.

However, if the angles are $\phi$ and $3\phi$, and the diameters cut the circle into sectors of angles $\phi, \phi, 3\phi, 3\phi$, then $2\phi + 6\phi = 2\pi$, so $8\phi = 2\pi$, and $\phi = \pi/4$. Then the angle between the diameters is $\phi = \pi/4$. We have $\tan \theta = \frac{2\sqrt{H^2 - AB}}{A+B}$. Alternatively, the diameters can make angles $\theta$ and $\pi/2 - \theta$ where $\theta = \pi/4$, so $\pi/2 - \pi/4 = \pi/4$. We can interpret this as the angle between the diameters is $\alpha$ and $\pi - \alpha$, where $\pi - \alpha = 3\alpha$, so $\pi = 4\alpha$, and $\alpha = \pi/4$. Or $\alpha = 3(\pi - \alpha)$, so $\alpha = 3\pi - 3\alpha$, and $4\alpha = 3\pi$, so $\alpha = 3\pi/4$. So the angle between the diameters is $\pi/4$.

Another way to think about this is that if the angle between the diameters is $\alpha$, then the other angle is $\pi/2 - \alpha$. We are given that one of these angles is three times the other. So either $\alpha = 3(\pi/2 - \alpha)$ or $\pi/2 - \alpha = 3\alpha$. In the first case, $\alpha = 3\pi/2 - 3\alpha$, so $4\alpha = 3\pi/2$, or $\alpha = 3\pi/8$. In the second case, $\pi/2 = 4\alpha$, so $\alpha = \pi/8$.

However, if we assume that the diameters are perpendicular, then the sectors are all equal.

Let the angle be $\theta$. Since one sector has area three times the other, we have $\theta = 3(\pi/2 - \theta)$, or $\pi/2 - \theta = 3\theta$. If $\theta = 3(\pi/2 - \theta)$, then $\theta = 3\pi/2 - 3\theta$, so $4\theta = 3\pi/2$, and $\theta = 3\pi/8$. If $\pi/2 - \theta = 3\theta$, then $\pi/2 = 4\theta$, so $\theta = \pi/8$.

The correct answer is $3a^2 - 10ab + 3b^2 = 0$. Let us try to derive this: $\tan \theta = \frac{2\sqrt{(a+b)^2 - ab}}{a+b} = \frac{2\sqrt{a^2+ab+b^2}}{a+b} = \tan(\pi/8)$ Then $\pi/8 = 22.5^{\circ}$. And $\tan(\pi/8) = \sqrt{2} - 1$.

$(\sqrt{2}-1)(a+b) = 2\sqrt{a^2+ab+b^2}$ $(2+1-2\sqrt{2})(a^2+2ab+b^2) = 4(a^2+ab+b^2)$ $(3-2\sqrt{2})(a^2+2ab+b^2) = 4(a^2+ab+b^2)$ $(3-2\sqrt{2})a^2 + (6-4\sqrt{2})ab + (3-2\sqrt{2})b^2 = 4a^2+4ab+4b^2$ $0 = (1+2\sqrt{2})a^2 + (4\sqrt{2}-2)ab + (1+2\sqrt{2})b^2$

Consider $\tan(3\pi/8) = \sqrt{2}+1$. $(\sqrt{2}+1)(a+b) = 2\sqrt{a^2+ab+b^2}$ $(2+1+2\sqrt{2})(a^2+2ab+b^2) = 4(a^2+ab+b^2)$ $(3+2\sqrt{2})(a^2+2ab+b^2) = 4(a^2+ab+b^2)$ $(3+2\sqrt{2})a^2 + (6+4\sqrt{2})ab + (3+2\sqrt{2})b^2 = 4a^2+4ab+4b^2$ $0 = (1-2\sqrt{2})a^2 + (2-4\sqrt{2})ab + (1-2\sqrt{2})b^2$

The correct answer must have been derived using $\tan \theta = \frac{2\sqrt{H^2 - AB}}{A+B}$.

$3a^2 - 10ab + 3b^2 = 0 \implies (3a-b)(a-3b)=0$

So $b = 3a$ or $a = 3b$. If $b=3a$, then $ax^2 + 2(a+3a)xy + 3ay^2 = 0$ which means $ax^2 + 8axy + 3ay^2 = 0$, so $x^2+8xy+3y^2 = 0$. Then $\tan \theta = \frac{2\sqrt{16-3}}{4} = \frac{\sqrt{13}}{2}$. If $a=3b$, then $3bx^2 + 2(3b+b)xy + by^2 = 0$ which means $3x^2 + 8xy + y^2 = 0$. Then $\tan \theta = \frac{2\sqrt{16-3}}{4} = \frac{\sqrt{13}}{2}$.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when squaring and simplifying expressions, especially when dealing with square roots and absolute values.
• Angle Interpretation: Remember that the $\tan^{-1}$ function returns an angle in the range $(-\pi/2, \pi/2)$. You may need to adjust the angle to find the correct angle between the lines.
• Checking the Answer: After obtaining a result, substitute it back into the original equation or conditions to verify its correctness.

Summary

The problem involves finding the relationship between the coefficients of a pair of lines given that they divide a circle into sectors with a specific area ratio. By using the formula for the angle between a pair of lines and the area relationship between the sectors, we set up an equation and simplified it to find the required relationship: $3a^2 + 2ab + 3b^2 = 0$. After reviewing the options given, it seems there might be an error in the problem statement or options. The correct derived equation is $3a^2 + 2ab + 3b^2 = 0$. However, the provided answer is $3a^2 - 10ab + 3b^2 = 0$.

The final answer is \boxed{3a^2 + 2ab + 3b^2 = 0}. The closest option is (D), but the sign of the $ab$ term differs.