Question

The locus of the foot of the perpendicular, from the origin to chords of the circle x² + y² – 4x – 6y – 3 = 0 which subtend a right angle at the origin, is –

• A. 2(x² + y²) – 4x – 6y – 3 = 0
• B. x² + y² + 4x + 6y + 3 = 0
• C. 2(x² + y²) +4x + 6y – 3 = 0
• D. None of these

Answer 1

Answer: (A)

2(x² + y²) – 4x – 6y – 3 = 0

Answer 2

The equation to one such chord, on which the foot of the perpendicular from the origin is

(x₁, y₁) is xx₁ + yy₁ = $x_{1}^{2}$ + $y_{1}^{2}$. Using this Homogenise

x² + y² – 4x – 6y – 3 = 0 .

This gives (x² + y²) ($x_{1}^{2}$ + $y_{1}^{2}$)² – 2(2x + 3y) (xx₁ + yy₁)

(x₁² + y₁²) – 3 (xx₁ + yy₁)² = 0. These two lines are at right angles.

\ 2($x_{1}^{2}$+$y_{1}^{2}$)² – 2(2x₁ + 3y₁) ($x_{1}^{2}$ + $y_{1}^{2}$) – 3 ($x_{1}^{2}$ + $y_{1}^{2}$) = 0

$x_{1}^{2}$ + $y_{1}^{2}$ ¹ 0 and hence locus of

(x₁, y₁) is 2(x² + y²) – 2(2x + 3y) – 3 = 0.