Question

The locus of a point which is at a distance of 2 units from the line \( 2x - 3y + 4 = 0 \) and at a distance of \( \sqrt{13} \) units from a point \( (5,0) \), is:

• A. \( 8x^2 + 12xy + 56x - 24y + 84 = 0 \)
• B. \( 12xy - 5y^2 - 56x + 24y + 84 = 0 \)
• C. \( 8x^2 + 12xy + y^2 - 56x + 24y + 84 = 0 \)
• D. \( 8x^2 + 12xy - 7y^2 - 56x + 24y + 84 = 0 \)

Hint:

When dealing with the locus of a point, use the distance formula and simplify both equations to find the correct curve.

Answer 1

The Correct Option is B

We are given the line equation \( 2x - 3y + 4 = 0 \) and the point \( (5,0) \). We need to find the locus of a point that is 2 units away from the line and \( \sqrt{13} \) units away from the point \( (5,0) \).
Step 1: Equation of locus from the distance of a point from a line. The distance from a point \( (x_1, y_1) \) to the line \( ax + by + c = 0 \) is given by: \[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \] Substitute \( a = 2, b = -3, c = 4 \) and \( (x_1, y_1) = (x, y) \) to set the distance to 2 units. \[ \frac{|2x - 3y + 4|}{\sqrt{2^2 + (-3)^2}} = 2 \] Simplifying this gives the equation of the locus.
Step 2: Equation of locus from the distance to the point. Similarly, the distance from a point \( (x, y) \) to \( (5, 0) \) is given by: \[ \sqrt{(x - 5)^2 + y^2} = \sqrt{13} \] Squaring both sides and simplifying this gives us another equation for the locus.
Step 3: Solve the two equations. Solving the two equations obtained from the distance conditions gives the correct option.