Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to : [2024]

Question

Four distinct points (2k, 3k), (1, 0), (0, 1) and (0, 0) lie on a circle for k equal to : [2024]

Smart Achievers · Accepted Answer

(2)

Let the equation of circle be

${(x - x_{1})}^{2} + {(y - y_{1})}^{2} = r^{2}$ ...(i)

Put x = 0, y = 0 in (i), we get

$\Rightarrow x_{1}^{2} + y_{1}^{2} = r^{2}$ ... (ii)

Put x = 0, y = 1 in (i), we get

$\Rightarrow x_{1}^{2} + {(1 - y_{1})}^{2} = r^{2}$ ... (iii)

From (ii) and (iii), we get

$y_{1}^{2} = {(1 - y_{1})}^{2} \Rightarrow y_{1}^{2} = 1 + y_{1}^{2} - 2 y_{1} \Rightarrow y_{1} = \frac{1}{2}$

Put x = 1, y = 0 in (i), we get

${(1 - x_{1})}^{2} + y_{1}^{2} = r^{2}$ ... (iv)

From (ii) and (iv), we get

${(1 - x_{1})}^{2} = x_{1}^{2} \Rightarrow 1 + x_{1}^{2} - 2 x_{1} = x_{1}^{2} \Rightarrow x_{1} = \frac{1}{2}$

$∴$ From (ii), we get $r = \frac{1}{\sqrt{2}}$

Putting $x_{1} = y_{1} = \frac{1}{2}$ and $r = \frac{1}{\sqrt{2}}$ in (i), we get

${(x - \frac{1}{2})}^{2} + {(y - \frac{1}{2})}^{2} = \frac{1}{2}$

Point (2k, 3k) also satisfies the equation of circle.

${(2 k - \frac{1}{2})}^{2} + {(3 k - \frac{1}{2})}^{2} = \frac{1}{2}$

$\Rightarrow 4 k^{2} + \frac{1}{4} - 2 k + 9 k^{2} + \frac{1}{4} - 3 k = \frac{1}{2}$

$\Rightarrow 13 k^{2} = 5 k \Rightarrow k = \frac{5}{13}$

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