Equation of Tangent in Slope Form
Trending Questions
Q. The equations of the tangents to the circle x2+y2 - 4x - 6y - 12 = 0 and parallel to 4x - 3y = 1 are
- 4x + 3y + 14 = 0, 4x + 3y + 16 = 0
- 4x – 3y – 24 = 0, 4x – 3y + 26 = 0
- x – y – 14 = 0, x – y + 16 = 0
- 4x – 3y + 34 = 0, 4x – 3y + 16 = 0
Q. Let tangents are drawn at two points of the circle (x−7)2+(y+1)2=25. If the point of intersection of both the tangents is origin, then the angle between them (in degrees) is
Q. The line lx + my + n = 0 will be a tangent to the circle x2+y2=a2 iff
- n2(l2+m2)=a2
- a2(l2+m2)=n2
- n(l + m) = a
- a(l + m) = n
Q. The equations of the tangents to the circle x2+y2−6x+4y=12 which are parallel to the straight line 4x + 3y + 5 = 0, are
- 3x - 4y + 19 = 0, 3x - 4y + 31 = 0
- 3x - 4y - 19 = 0, 3x - 4y + 31 = 0
- 4x + 3y - 19 = 0, 4x + 3y + 31 = 0
- 4x + 3y + 19 = 0, 4x + 3y - 31 = 0
Q.
The equation of tangent with slope m of the circle x2+y2=a2 are given by y=mx±m√1+a2
True
False
Q. The equations of the tangents drawn from the origin to the circle x2+y2−2rx−2hy+h2=0 are
- x=0
- y=0
- (h2−r2)x−2rhy=0
- (h2−r2)x+2rhy=0
Q.
A tangent PT is drawn to the circle x2+y2=4 at the point P(√3, 1). A straight line L, perpendicular to PT is a tangent to the circle (x−3)2+y2=1
A possible equation of L is
x−√3y=1
x+√3y=1
x−√3y=−1
x+√3y=5
Q. A tangent PT is drawn to the circle x2+y2=4 at the point P(√3, 1). A straight line L, perpendicular to PT is a tangent to the circle (x−3)2+y2=1
A common tangent of the two circles is
A common tangent of the two circles is
- x=4
- y=2
- x+√3y=4
- x+2√2y=6
Q. Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, is
- 2√3y=−x−12
- 2√3y=12x+1
- √3y=x+3
- √3y=3x+1
Q. The area (in sq. units) of the smaller of the two circles that touch the parabola, y2=4x at the point (1, 2) and the x-axis is :
- 4π(3+√2)
- 8π(2−√2)
- 8π(3−2√2)
- 4π(2−√2)