If the lines 2x - 3y = 5 and 3x - 4y = 7 are two diameters of a circle of radius 7, then the equation of the circle is
x2 + y2 + 2x - 4y - 47 = 0
x2 + y2 = 49
x2 + y2 - 2x + 2y - 47 = 0
x2 + y2 = 17
C.
x2 + y2 - 2x + 2y - 47 = 0
Since, the lines 2x - 3y = 5 and 3x - 4y = 7 are the diameters of a circle. Therefore, the point of intersection is the centre of the circle. On solving the given equations, we get x = 1 and y = - 1 ie, the centre of the circle.
:. Required equation of circle is
(x - 1)2 + (y + 1)2 = 72
⇒ x2 + y2 - 2x + 2y + 2 = 49
⇒ x2 + y2 - 2x + 2y - 47 = 0
If is the angle between the tangents from(- 1, 0) to the circle x2 + y2 - 5x + 4y - 2 = 0, then is equal to
If 2x + 3y + 12 = 0 and x - y + 4 = 0 are conjugate with respect to the parabola y = 8x, then is equal to
2
- 2
3
- 3
For an ellipse with eccentricity the centre is at the origin. If one directrix is x = 4, then the equation of the ellipse is
The area (in square unit) of the circle which touches the lines 4x + 3y = 15 and 4x + 3y = 5 is
4
The equations of the circle which pass through the origin and makes intercepts of length 4 and 8 on the x and y-axes respectively are