Question

Let \( A(2, 3), B(1, -1) \) be two points. If \( P \) is a variable point such that \( \angle APB = 90^\circ \), then the locus of \( P \) is

• A. \( x^2 + y^2 - x - 4y + 1 = 0 \)
• B. \( x^2 + y^2 + x + 4y - 1 = 0 \)
• C. \( x^2 + y^2 - x + 4y - 1 = 0 \)
• D. \( x^2 + y^2 + x - 4y + 1 = 0 \)

Hint:

The locus of a point forming a right angle with two fixed points lies on a circle with the line segment joining the fixed points as the diameter. Use the midpoint formula for the center and the distance formula for the radius to find the equation of the circle.

Answer 1

The Correct Option is A

Step 1: Knowing \( \angle APB = 90^\circ \), the points \( A(2, 3) \) and \( B(1, -1) \) with \( P(x, y) \) form a right triangle where \( AP \) and \( BP \) are perpendicular. Step 2: Use the circle theorem that states the angle in a semi-circle is a right angle. Thus, \( P \) lies on the circle with diameter \( AB \). 
Step 3: Calculate the equation of the circle having diameter \( AB \). - The midpoint \( M \) of \( AB \) is the average of the coordinates of \( A \) and \( B \): \[ M = \left( \frac{2+1}{2}, \frac{3 + (-1)}{2} \right) = \left( \frac{3}{2}, 1 \right). \] - The radius is the distance from \( M \) to \( A \) (or \( M \) to \( B \), which will be the same): \[ r = \sqrt{\left( 2 - \frac{3}{2} \right)^2 + \left( 3 - 1 \right)^2} = \sqrt{\left( \frac{1}{2} \right)^2 + 2^2} = \sqrt{\frac{1}{4} + 4} = \sqrt{\frac{17}{4}} = \frac{\sqrt{17}}{2}. \] - The equation of the circle with center \( M \left( \frac{3}{2}, 1 \right) \) and radius \( r = \frac{\sqrt{17}}{2} \) is: \[ \left( x - \frac{3}{2} \right)^2 + (y - 1)^2 = \frac{17}{4}. \] Step 4: Expand and simplify the equation of the circle: \[ \left( x - \frac{3}{2} \right)^2 + (y - 1)^2 = \frac{17}{4}. \] Expanding: \[ \left( x^2 - 3x + \frac{9}{4} \right) + \left( y^2 - 2y + 1 \right) = \frac{17}{4}. \] Multiply through by 4 to eliminate the fractions: \[ 4x^2 - 12x + 9 + 4y^2 - 8y + 4 = 17. \] Simplifying further: \[ 4x^2 + 4y^2 - 12x - 8y + 13 = 17 \quad \Rightarrow \quad 4x^2 + 4y^2 - 12x - 8y - 4 = 0. \] Finally, divide through by 4 to simplify: \[ x^2 + y^2 - 3x - 2y - 1 = 0. \]