Question

Let $ A(4, 3), B(2, 5) $ be two points. If $ P $ is a variable point on the same side of the origin as that of line $ AB $ and at most 5 units from the midpoint of $ AB $, then the locus of $ P $ is:

• A. \( x^2 + y^2 - 6x - 8y = 0 \)
• B. \( x^2 + y^2 - 6x - 8y \le 0, \quad x + y - 7<0 \)
• C. \( x^2 + y^2 + 6x + 8y - 25 = 0, \quad x + y - 7 \le 0 \)
• D. \( x^2 + y^2 - 6x + 8y \ge 0, \quad x + y - 7<0 \)

Hint:

Use midpoint formula and circle equation for locus; use line equation and sign test for side of point relative to line.

Answer 1

The Correct Option is B

1. Midpoint of \( AB \) is: \[ M = \left( \frac{4+2}{2}, \frac{3+5}{2} \right) = (3, 4) \] 2. The locus of points at most 5 units from \( M \) is a circle: \[ (x - 3)^2 + (y - 4)^2 \le 25 \] Expanding: \[ x^2 - 6x + 9 + y^2 - 8y + 16 \le 25 \Rightarrow x^2 + y^2 - 6x - 8y \le 0 \] 3. The line \( AB \) has equation: \[ \frac{y - 3}{5 - 3} = \frac{x - 4}{2 - 4} \Rightarrow \frac{y - 3}{2} = \frac{x - 4}{-2} \Rightarrow x + y - 7 = 0 \] 4. \( P \) lies on the same side of the origin relative to \( AB \): \[ \text{At origin } (0,0), \quad 0 + 0 -7 = -7<0 \] So the region is: \[ x + y - 7<0 \]