The equation of the diameter of the circle x2+y2+2 x-4 y-11=0 which bisects the chords intercepted on the line 2 x-y+3=0 is:

Question

The equation of the diameter of the circle x2+y2+2 x-4 y-11=0 which bisects the chords intercepted on the line 2 x-y+3=0 is: (A)  x+y-7=0  (B)  2 x-y-5=0  (C) x+2 y-3=0  (D) none of these

Tardigrade Admin · Accepted Answer

The perpendicular distance from centre to the chord its bisects the chord.
Given equation of circle is

x^{2} + y^{2} + 2 x - 4 y - 11 = 0

∴

Centre is,

(- 1, 2)

.
Since, the diameter of the circle is perpendicular to the chord of the circle. Hence, the equation of diameter is

x + 2 y + λ = 0 \dots

..(i)
Its passes through the centre

(- 1, 2)

of the circle

\Rightarrow - 1 + 4 + λ = 0

\Rightarrow λ = - 3

Putting the value of

λ

in Eq. (i), we get

x + 2 y - 3 = 0

Q. The equation of the diameter of the circle $x^{2} + y^{2} + 2 x - 4 y - 11 = 0$ which bisects the chords intercepted on the line $2 x - y + 3 = 0$ is:

$x + y - 7 = 0$

$2 x - y - 5 = 0$

$x + 2 y - 3 = 0$

none of these

Q. The equation of the diameter of the circle x2+y2+2x−4y−11=0 which bisects the chords intercepted on the line 2x−y+3=0 is:

x+y−7=0

2x−y−5=0

x+2y−3=0

none of these

Q. The equation of the diameter of the circle $x^{2} + y^{2} + 2 x - 4 y - 11 = 0$ which bisects the chords intercepted on the line $2 x - y + 3 = 0$ is:

$x + y - 7 = 0$

$2 x - y - 5 = 0$

$x + 2 y - 3 = 0$