Question

If $x=3$ is the chord of contact of the circle $x^{2}+y^{2}=81$ , then the equation of the corresponding pair of tangents is

• A. $x^{2}-8y^{2}+54x+729=0$
• B. $x^{2}-8y^{2}-54x+729=0$
• C. $x^{2}-8y^{2}-54x-729=0$
• D. $x^{2}-8y^{2}=729$

Answer 1

Answer: (B)

Put $x=3$ in the equation of circle to get the point of contact, $y=\pm6\sqrt{2}$
Hence, equation of the pair of tangents is
$\left(3 x + 6 \sqrt{2} y - 81\right)\left(3 x - 6 \sqrt{2} y - 81\right)=0$
$\left(x - 27\right)^{2}-8y^{2}=0$
$x^{2}-54x-8y^{2}+729=0$