Question

Let circle \( C \) be the image of

\[ x^2 + y^2 - 2x + 4y - 4 = 0 \]

in the line

\[ 2x - 3y + 5 = 0 \]

and \( A \) be the point on \( C \) such that \( OA \) is parallel to the x-axis and \( A \) lies on the right-hand side of the centre \( O \) of \( C \).

If \( B(\alpha, \beta) \), with \( \beta < 4 \), lies on \( C \) such that the length of the arc \( AB \) is \( \frac{1}{6} \) of the perimeter of \( C \), then \( \beta - \sqrt{3}\alpha \) is equal to:

• A. 3
• B. \( 3 + \sqrt{3} \)
• C. \( 4 - \sqrt{3} \)
• D. 4

Hint:

To find the distance between two points on a circle, use the arc length formula. The angle subtended by the arc at the center is related to the arc length and the radius. Use the parametric equations of the circle to find the coordinates of points on the circle.

Answer 1

The Correct Option is D

Step 1: Find the center and radius of the original circle. The equation of the original circle is \( x^2 + y^2 - 2x + 4y - 4 = 0 \). 
Completing the square, we have \( (x^2 - 2x + 1) + (y^2 + 4y + 4) = 4 + 1 + 4 \), which simplifies to \( (x-1)^2 + (y+2)^2 = 9 \). 
The center of this circle is \( (1, -2) \) and the radius is \( r = 3 \).
Step 2: Find the center of the image circle. Since the line \( 2x - 3y + 5 = 0 \) acts as a mirror, the radius of the image circle C is the same, \( r = 3 \). The center of the image circle is the reflection of \( (1, -2) \) in the line \( 2x - 3y + 5 = 0 \). Let the center of C be \( (h, k) \).
The midpoint of \( (1, -2) \) and \( (h, k) \) lies on the line \( 2x - 3y + 5 = 0 \). The midpoint is \( \left( \frac{h+1}{2}, \frac{k-2}{2} \right) \), so \[ 2\left( \frac{h+1}{2} \right) - 3\left( \frac{k-2}{2} \right) + 5 = 0 \]\[\implies 2(h+1) - 3(k-2) + 10 = 0 \implies 2h + 2 - 3k + 6 + 10 = 0 \implies 2h - 3k + 18 = 0.\]
The line joining \( (1, -2) \) and \( (h, k) \) is perpendicular to the line \( 2x - 3y + 5 = 0 \), so the slope of the line joining \( (1, -2) \) and \( (h, k) \) is \( -\frac{3}{2} \).
\[ \frac{k - (-2)}{h - 1} = -\frac{3}{2} \implies 2(k+2) = -3(h-1) \implies 2k + 4 = -3h + 3 \implies 3h + 2k + 1 = 0. \] Now we solve the system of equations: \[ 2h - 3k + 18 = 0 \] \[ 3h + 2k + 1 = 0 \] Multiply the first equation by 2 and the second equation by 3: \[ 4h - 6k + 36 = 0 \] \[ 9h + 6k + 3 = 0 \] Adding the two equations gives \( 13h + 39 = 0 \), so \( h = -3 \). Substituting into the second equation: \[ 3(-3) + 2k + 1 = 0 \implies -9 + 2k + 1 = 0 \implies 2k = 8 \implies k = 4. \] The center of the image circle C is \( (-3, 4) \). 
Step 3: Find the coordinates of point A. Point A lies on the circle C and OA is parallel to the x-axis, so A has y-coordinate 4. 
Also A lies on the right hand side of the center O of C, meaning A has x>-3. Since A lies on C, \( (x+3)^2 + (y-4)^2 = 9 \implies (x+3)^2 + (4-4)^2 = 9 \implies (x+3)^2 = 9 \implies x+3 = \pm 3 \). 
Then \( x = -3 \pm 3 \). So \( x = 0 \) or \( x = -6 \). Since we need \( x>-3 \), we must have \( x = 0 \). So A is \( (0, 4) \).
Step 4: Find the coordinates of point B.
The arc length AB is \( \frac{1}{6} \) of the perimeter of C. The perimeter is \( 2 \pi r = 2 \pi (3) = 6\pi \). 
The arc length AB is \( \frac{1}{6}(6\pi) = \pi \). 
If \( (\alpha, \beta) \) are the coordinates of B, then we have \( (\alpha + 3)^2 + (\beta - 4)^2 = 9 \).
We use the angle subtended at the center to find B. 
Let the angle be \(\theta\). The arc length is \( r\theta\), so \( \pi = 3 \theta \) and thus \( \theta = \frac{\pi}{3} \). If A is at \( 0 \) on the circle's parametrization, B is at \( \pm \pi/3 \). Let \( x = -3 + 3\cos(\theta), y = 4 + 3\sin(\theta)\).
Then we have \(B = (-3 + 3 \cos( \frac{\pi}{3} ), 4 + 3\sin( \frac{\pi}{3} ) ) = (-3 + \frac{3}{2}, 4 + \frac{3\sqrt{3}}{2}) \) or \(B = (-3 + 3 \cos( -\frac{\pi}{3} ), 4 + 3\sin( -\frac{\pi}{3} ) ) = (-3 + \frac{3}{2}, 4 - \frac{3\sqrt{3}}{2})\).
Since \( \beta<4 \), we choose the second case. Thus, \(\alpha = -\frac{3}{2}, \beta = 4 - \frac{3\sqrt{3}}{2} \). 
Step 5: Compute \( \beta - \sqrt{3} \alpha \).
Then \( \beta - \sqrt{3} \alpha = 4 - \frac{3\sqrt{3}}{2} - \sqrt{3}(-\frac{3}{2}) = 4 - \frac{3\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 4 \). 
Final Answer: The value of \( \beta - \sqrt{3}\alpha \) is 4. The correct answer is (4).