Question

If $A(2, 4)$ and $B(6, 10)$ are two fixed points and if a point $P$ moves so that $\angle APB$ is always a right angle, then the locus of $P$ is:

• A. $x^2 + y^2 + 8x + 14y + 52 = 0$
• B. $x^2 + y^2 - 8x + 14y - 52 = 0$
• C. $x^2 + y^2 + 8x - 14y + 52 = 0$
• D. $x^2 + y^2 - 8x - 14y - 52 = 0$
• E. $x^2 + y^2 - 8x - 14y + 52 = 0$

Hint:

Alternatively, you can use the slope condition: $m_{AP} \times m_{BP} = -1$. This results in $\frac{y-4}{x-2} \times \frac{y-10}{x-6} = -1$, which leads to the same diameter form equation.

Answer 1

Answer: (E)

Concept: A point $P(x, y)$ that maintains a $90^\circ$ angle with two fixed points $A$ and $B$ describes a circle where $AB$ is the diameter. This is based on the geometric theorem that the angle subtended by a diameter at any point on the circumference is a right angle.
• Equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$:
• $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.

Step 1: Set up the diameter form equation.
Given $A(x_1, y_1) = (2, 4)$ and $B(x_2, y_2) = (6, 10)$: \[ (x - 2)(x - 6) + (y - 4)(y - 10) = 0 \]

Step 2: Expand the algebraic terms.
$x$ terms: $(x - 2)(x - 6) = x^2 - 6x - 2x + 12 = x^2 - 8x + 12$.
$y$ terms: $(y - 4)(y - 10) = y^2 - 10y - 4y + 40 = y^2 - 14y + 40$.

Step 3: Combine into the general form.
Combine the two expressions: \[ (x^2 - 8x + 12) + (y^2 - 14y + 40) = 0 \] \[ x^2 + y^2 - 8x - 14y + (12 + 40) = 0 \] \[ x^2 + y^2 - 8x - 14y + 52 = 0 \]