The equation of a circle in which the chord joining the points $(1, 2)$ and $(2, - 1)$ subtends an angle of $\frac{π}{4}$ at any point on the circumference is

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$x^{2} + y^{2} - 6 x - 2 y + 5 = 0$

$x^{2} + y^{2} - 2 x - 4 y + 4 = 0$

$x^{2} + y^{2} + 6 x + 2 y - 15 = 0$

$x^{2} + y^{2} - 5 = 0$

answer is A, B.

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Detailed Solution

Slope of $P A = (m_{1}) = \frac{y - 2}{x - 1}$

Slope of $P B = m_{2} = \frac{y + 1}{x - 2}$

$θ = \frac{π}{4}$

$\tan θ = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

$\tan \frac{π}{4} = | \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} |$

$1 + m_{1} m_{2} = | m_{1} - m_{2} |$

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

$(x - 1) (x - 2) + (y - 2) (y + 1) = | (y - 2) (x - 2) - (y + 1) (x - 1) |$

$x^{2} + y^{2} - 3 x - y = \pm | x y - 2 y - 2 x + 4 - x y + y - x + 1 |$

$x^{2} + y^{2} - 3 x - y = \pm |- y - 3 x + 5|$

$(i) x^{2} + y^{2} - 3 x - y = - y - 3 x + 5$

$x^{2} + y^{2} - 5 = 0$

$(i i) x^{2} + y^{2} - 3 x - y = - [- y - 3 x + 5]$

$x^{2} + y^{2} - 3 x - y = y + 3 x - 5$

$x^{2} + y^{2} - 6 x - 2 y + 5 = 0$

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