Question

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

• A. x^2 + y^2 - 5 = 0
• B. x^2 + y^2 - 6x - 2y + 5 = 0
• C. x^2 + y^2 + 6x + 2y - 15 = 0
• D. x^2 + y^2 + 7x - 2y + 14 = 0

Answer 1

Answer: (A), (B)

Both A and B are correct.

Answer 2

The general equation of a circle where a chord joining points $A(x_1, y_1)$ and $B(x_2, y_2)$ subtends an angle $\theta$ at any point $P(x, y)$ on the circumference is given by: $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = \pm \cot \theta [(x-x_1)(y-y_2) - (x-x_2)(y-y_1)]$ Given points are $A(1, 2)$ and $B(2, -1)$, and the angle $\theta = \frac{\pi}{4}$. Thus, $\cot \theta = \cot \frac{\pi}{4} = 1$.

Substituting the coordinates of A and B: $(x-1)(x-2) + (y-2)(y-(-1)) = \pm 1 \cdot [(x-1)(y-(-1)) - (x-2)(y-2)]$ $(x^2 - 3x + 2) + (y^2 - y - 2) = \pm [(x-1)(y+1) - (x-2)(y-2)]$ $x^2 + y^2 - 3x - y = \pm [ (xy + x - y - 1) - (xy - 2x - 2y + 4) ]$ $x^2 + y^2 - 3x - y = \pm [ xy + x - y - 1 - xy + 2x + 2y - 4 ]$ $x^2 + y^2 - 3x - y = \pm (3x + y - 5)$

Case 1: Using the '+' sign $x^2 + y^2 - 3x - y = 3x + y - 5$ $x^2 + y^2 - 6x - 2y + 5 = 0$ This matches option (B).

Case 2: Using the '-' sign $x^2 + y^2 - 3x - y = -(3x + y - 5)$ $x^2 + y^2 - 3x - y = -3x - y + 5$ $x^2 + y^2 - 5 = 0$ This matches option (A).

Both equations (A) and (B) satisfy the given condition. The question asks for "The equation of a circle", and both derived equations are valid. Therefore, both options are correct.