If x=3 is the chord of contact of the circle x2+y2=81 , then the equation of the corresponding pair of tangents is

Question

If x=3 is the chord of contact of the circle x2+y2=81 , then the equation of the corresponding pair of tangents is (A) x2-8y2+54x+729=0 (B) x2-8y2-54x+729=0 (C) x2-8y2-54x-729=0 (D) x2-8y2=729

Tardigrade Admin · Accepted Answer

Put x=3 in the equation of circle to get the point of contact, y=±6√2 Hence, equation of the pair of tangents is (3 x + 6 √2 y - 81)(3 x - 6 √2 y - 81)=0 (x - 27)2-8y2=0 x2-54x-8y2+729=0 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-tw76w2svwahnkj9u.jpg" />

Q. 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

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

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

$x^{2} - 8 y^{2} - 54 x - 729 = 0$

$x^{2} - 8 y^{2} = 729$

Put $x = 3$ in the equation of circle to get the point of contact, $y = \pm 62$
Hence, equation of the pair of tangents is
$(3 x + 62 y - 81) (3 x - 62 y - 81) = 0$
$(x - 27)^{2} - 8 y^{2} = 0$
$x^{2} - 54 x - 8 y^{2} + 729 = 0$

Q. If x=3 is the chord of contact of the circle x2+y2=81 , then the equation of the corresponding pair of tangents is

x2−8y2+54x+729=0

x2−8y2−54x+729=0

x2−8y2−54x−729=0

x2−8y2=729