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Mathematics
The equation of the tangents to the circle x2 + y2 - 6x + 4y = 12, which are parallel to the straight line 4x + 3y + 5 = 0, are
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Q. The equation of the tangents to the circle $x^2 + y^2 - 6x + 4y = 12$, which are parallel to the straight line $4x + 3y + 5 = 0,$ are
Conic Sections
A
$3x - 4y - 19 = 0, 3x - 4y + 31 = 0 $
B
$4x + 3y - 19 = 0, Ax + 3y + 31 = 0 $
C
$4x + 3y + 19 = 0, 4x + 3y - 31 = 0 $
D
$3x - 4y + 19 = 0, 3x - 4y + 31 = 0. $
Solution:
Correct answer is (c) $4x + 3y + 19 = 0, 4x + 3y - 31 = 0 $