Question

If \(P\) is a variable point which is at a distance of 2 units from the line \(2x - 3y + 1 = 0\) and \(\sqrt{13}\) units from the point (5, 6), then the equation of the locus of \(P\) is

• A. \(4x^2 + 12xy - 5y^2 - 44x - 42y + 245 = 0\)
• B. \(12xy - 5y^2 - 44x - 42y + 243 = 0\)
• C. \(8x^2 + 12xy - 5y^2 - 44x - 42y + 243 = 0\)
• D. \(12xy - 13y^2 - 44x - 42y + 245 = 0\)

Hint:

Convert distance conditions into equations, then eliminate to derive the locus.

Answer 1

The Correct Option is B

Let \(P(x, y)\) satisfy both: \[ \text{Distance from line } \dfrac{|2x - 3y + 1|}{\sqrt{13}} = 2 \Rightarrow (2x - 3y + 1)^2 = 4 \cdot 13 = 52 \] Also, \[ \text{Distance from (5, 6) } = \sqrt{(x - 5)^2 + (y - 6)^2} = \sqrt{13} \Rightarrow (x - 5)^2 + (y - 6)^2 = 13 \] Expand both equations and subtract one from the other to eliminate constants. The resulting equation is the required locus.