The equation of the pair of straight lines parallel to x-axis and touching the circle x2 + y2- 6x - 4y -12 = 0 is
y2 - 4y - 21 = 0
y2 + 4y - 21 = 0
y2 - 4y + 21 = 0
y2 + 4y + 21 = 0
A.
y2 - 4y - 21 = 0
Let the lines be y = m1x + c1 and y = m2x + c2.
Since, pair of straight lines are parallel to x-axis
and the lines will be y = c1 and y = c2.
Given circle is x2 + y2 - 6x - 4y - 12 = 0
Centre (3, 2) and radius = 5
Here, the perpendicular drawn from centre to the lines are CP and CP'.
Hence, the lines are
y - 7 = 0, y + 3 = 0 i.e., (y - 7)(0 + 3) = 0
Pair of straight lines is y2 - 4y - 21 = 0
If the foot of the perpendicular from the origin to a straight line is at the point (3 - 4). Then, the equation of the line is
3x - 4y = 25
3x - 4y + 25 = 0
4x + 3y - 25 = 0
4x - 3y + 25 = 0
The angle between lines joining origin and intersection points of line 2x + y = 1 and curve 3x2 + 4yx - 4x + 1= 0 is
The equation of the bisector of the acute angle between the lines 3x - 4y + 7 = 0 and 12x + 5y - 2 = 0 is
99x - 27y - 81 = 0
11x - 3y + 9 = 0
21x + 77y - 101 = 0
21x + 77y + 101 = 0
The value of k so that x2 + y2 + kx + 4y + 2 = 0 and 2(x2 + y) - 4x - 3y + k = 0 cut orthogonally is
The two lines x = my + n, z = py + q and x = m'y + n', z =p'y + q' are perpendicular to each other, if
mm' + pp' = 1
mm' + pp' = - 1