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 $x$ -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 $(1/6)^{\text{th}}$ of the perimeter of $C$, then $\beta - \sqrt{3} \alpha$ is equal to

Step 1: Find the center and radius of the original circle

First, we need to find the center and radius of the original circle. We can rewrite the equation of the circle in the standard form $$(x-h)^2 + (y-k)^2 = r^2$$ by completing the square.

Step 2: Complete the square for the original circle

Completing the square for the $$x$$ terms gives $$(x-1)^2$$ and for the $$y$$ terms gives $$(y+2)^2$$ . We add and subtract the constants to maintain equality. This shows the original circle has center $$P(1, -2)$$ and radius $$r = 3$$ .

Step 3: Find the image of the center of the circle

The image of the circle $$C$$ in the line $$2x - 3y + 5 = 0$$ will have the same radius as the original circle. Its center O(h, k) will be the reflection of the original center $$P(1, -2)$$ across the line $$2x - 3y + 5 = 0$$ .

Step 4: Calculate the reflected center O

Using the formula for the reflection of a point $$(x_1, y_1)$$ across the line $$ax + by + c = 0$$ , which is $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$$ , we find the coordinates of the new center $$O$$ . The center of circle $$C$$ is $$O(-3, 4)$$ and its radius is $$r=3$$ .

Step 5: Determine point A

Point $$A$$ is on circle $$C$$ , $$OA$$ is parallel to the $$x$$ -axis, and $$A$$ is to the right of the center $$O(-3, 4)$$ . This means $$A$$ has the same $$y$$ -coordinate as $$O$$ and its $$x$$ -coordinate is $$x_O + r$$ . So, $$A = (-3+3, 4) = (0, 4)$$ .

Step 6: Determine point B

The length of arc $$AB$$ is $$1/6$$ th of the perimeter of $$C$$ . This implies that the angle subtended by the arc $$AB$$ at the center $$O$$ is $$1/6$$ th of the total angle in a circle, which is $$2\pi$$ radians or $$360^\circ$$ . So, $$\angle AOB = 60^\circ$$ .

Step 7: Calculate coordinates of B and the final expression

Point $$A$$ is at $$(0, 4)$$ and the center $$O$$ is at $$(-3, 4)$$ . The vector $$\vec{OA}$$ is $$(3, 0)$$ , which lies along the positive $$x$$ -axis relative to $$O$$ . Since $$\angle AOB = 60^\circ$$ and $$\beta < 4$$ (meaning $$B$$ is below the line $$y=4$$ ), point $$B$$ is obtained by rotating $$A$$ by $$-60^\circ$$ around $$O$$ . The coordinates of $$B$$ are $$(x_O + r\cos(-60^\circ), y_O + r\sin(-60^\circ))$$ . Substituting the values, we get $$\alpha = -3 + 3(1/2) = -3/2$$ and $$\beta = 4 + 3(-\sqrt{3}/2) = 4 - 3\sqrt{3}/2$$ . Finally, we calculate $$\beta - \sqrt{3}\alpha = (4 - 3\sqrt{3}/2) - \sqrt{3}(-3/2) = 4 - 3\sqrt{3}/2 + 3\sqrt{3}/2 = 4$$ .

More from this chapter

• Find the equation of the circle passing through points (1,2), (3,4), and (5,2).
• Determine the equation of the circle with radius 4 and Centre (-2, 3).
• Find the equation of the hyperbola where foci are (0, +-6) and the length of the latus rectum is 36.
• Find the equation of the hyperbola where foci are (0, ± 12) and the length of the latus rectum is 36
• Find the equation of the circle with centre (-3,2) and radius 10.
• Find the equation of the circle with centre (-3,2) and radius 9.
• Find the equation of the circle with centre (-3,2) and radius 8.
• Find the equation of the circle with centre (-3,2) and radius 6.

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