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

The correct option is B $x^2-8y^2-54x+729=0$

The equation of chord of contanct is

$x=3.....\left(1\right)$

The equation of given circle is-

$x^2+y^2=81.....\left(2\right)$

Point of intersection of $\left(1\right)\&\left(2\right)$

$3^2+y^2=81$

$\Rightarrow y^2=72$

$\Rightarrow y=\pm 6\sqrt{2}$

$\therefore$ Point of intersection are $(3,6\sqrt{2})$ and $(3,-6\sqrt{2})$

Now, equation of tangent to the circle $x^2-y^2=81$ at $(3,6\sqrt{2})$ is-

$x\times 3+y\times 6\sqrt{2}=81$

$\Rightarrow x+2\sqrt{2}y-27=0.....\left(3\right)$

Now, equation of tangent to the circle $x^2+y^2=81$ at $(3,-6\sqrt{2})$ is-

$x\times 3+y\times (-6\sqrt{2})=81$

$\Rightarrow x-2\sqrt{2}y-27=0.....\left(4\right)$

Equation $\left(3\right)\&\left(4\right)$ are the equation of tangents to the circle $x^2+y^2=81$.

Now, joint equation of the pair of tangent is-

$(x+2\sqrt{2}y-27)(x-2\sqrt{2}y-27)=0$

${(x-27)}^2-{\left(2\sqrt{2}y\right)}^2=0$

$x^2+729-54x-8y^2=0$

$\Rightarrow x^2-8y^2-54x+729=0$

Hence the equation of the corresponding pair of tangents is $x^2-8y^2-54x+729=0$.