Let circle C be the image of 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 , with , lies on C such that the length of the arc AB is of the perimeter of C

Question

Let circle C be the image of $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ 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 $B (α, β)$ , with $β < 4$ , lies on C such that the length of the arc AB is ${(1 / 6)}^{th}$ of the perimeter of C, then $β - \sqrt{3} α$ is equal to [2025]

Smart Achievers · Accepted Answer

(3)

Centre of the circle $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ is (1, –2) and its radius $\sqrt{1 + 4 + 4} = 3$

Image of (1, –2) about 2x – 3y + 5 = 0 is

$\frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{- 2 (2 + 6 + 5)}{4 + 9} = - 2$

$∴$ x = – 3 and y = 4

Centre of circle C is (–3, 4).

Equation of Circle C is ${(x + 3)}^{2} + {(y - 4)}^{2} = 9$

$l$ $(arc A B) = \frac{1}{6} \times 2 π r \Rightarrow r θ = \frac{1}{6} \times 2 π r \Rightarrow θ = \frac{π}{3}$

As $(α, β)$ lies on circle C,

$\Rightarrow {(α + 3)}^{2} + {(β - 4)}^{2} = 9$ ... (i)

$\Rightarrow$ Coordinates of A are (0, 4) ( $∵$ A lies on right side of O and OA $| |$ x-axis)

Now, $A B^{2} = O A^{2} + O B^{2} - 2 O A \cdot O B \cdot \cos \frac{π}{3}$

$= 9 + 9 - 2 \times 3 \times 3 \times \frac{1}{2} = 9$

$\Rightarrow α^{2} + {(β - 4)}^{2} = 9$ ... (ii)

From (i) and (ii), we get $α = \frac{- 3}{2}$

${(β - 4)}^{2} = 9 - \frac{9}{4} \Rightarrow β = \frac{- 3 \sqrt{3}}{2} + 4 (∵ β < 4)$

Now, $β - \sqrt{3} α = \frac{- 3 \sqrt{3}}{2} + 4 - \sqrt{3} (\frac{- 3}{2}) = 4$ .

Solution

1 $3 + \sqrt{3}$

2 $4 - \sqrt{3}$

3 4

4 3

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Solution

1 3+3

2 4–3

3 4

4 3

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1 $3 + \sqrt{3}$

2 $4 - \sqrt{3}$