The equation of the diameter of the circle (x-2)2+(y+1)2=16 which bisects the chord cut off by the circle on the line x-2 y-3=0 is

Question

The equation of the diameter of the circle (x-2)2+(y+1)2=16 which bisects the chord cut off by the circle on the line x-2 y-3=0 is (A) x+2 y=0 (B) 2 x+y-3=0 (C) 3 x+2 y-4=0 (D) none

Tardigrade Admin · Accepted Answer

Required diameter is perp to given line. Hence y+1=-2(x-2) ⇒ 2 x+y-3=0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/45677730d12793e2c8d47b267b609400-.png" />

Q. The equation of the diameter of the circle $(x - 2)^{2} + (y + 1)^{2} = 16$ which bisects the chord cut off by the circle on the line $x - 2 y - 3 = 0$ is

$x + 2 y = 0$

$2 x + y - 3 = 0$

$3 x + 2 y - 4 = 0$

none

Required diameter is $⊥$ to given line.
Hence $y + 1 = - 2 (x - 2)$
$\Rightarrow 2 x + y - 3 = 0$

Q. The equation of the diameter of the circle (x−2)2+(y+1)2=16 which bisects the chord cut off by the circle on the line x−2y−3=0 is

x+2y=0

2x+y−3=0

3x+2y−4=0