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Mathematics is an important subject in SSC,DSSB & Other Exams. .

Q1. The equation of the tangent to the circle x^2+y^2=r^2 at (a, b) is ax+by-λ=0, where  is :
•  a^2
•  b^2
•  r^2
•  None of these
Solution
r^2

Q2.x=7 touches the circle x^2+y^2-4x-6y-12=0, then the coordinates of the point of contact are :
•  (7, 3):
•  (7, 4)
•  (7, 8)
•  (7, 2)
Solution
(7, 3)

Q3.  A circle with centre (a, b) passes through the origin. The equation of the tangent to the circle at the origin is :
•  ax-by=0
•  ax+by=0
•  bx-ay=0
•  ((a^2+b^2)/a^2 ,(a^2+b^2)/b^2 )
Solution
ax+by=0

Q4.  If the tangent at a point P(x,y) of a curve is perpendicular to the line that joins origin with the point P, then the curve is
•  Circle
•  Parabola
•  Ellipse
•  Straight line
Solution
Circle

Q5.The circle x^2+y^2-8x+4y+4=0 touches
•  x-axis only
•  y-axis only
•  Both x and y-axis
•  Does not touch any axis
Solution
Y-axis only

Q6. The condition that the line x cos⁡α+y sin⁡α=p may touch the circle x^2+y^2-5x+6y+15=0 is
•  p=a cos⁡α
•  p=a tan⁡α
• p^2=a^2
•  x^2+y^2-2x-6y+9=0
Solution
p^2=a^2

Q7.The equation of circle with centre (1, 2) and tangent x+y-5=0 is
•  x^2+y^2+2x-4y+6=0
•  x^2+y^2-2x-4y+3=0
•  x^2+y^2-2x+4y+8=0
•  x^2+y^2-2x-4y+8=0
Solution
x^2+y^2-2x-4y+3=0

Q8.The equation of tangent to the circle x^2+y^2=a^2 parallel to y=mx+c is
•  y=mx±√(1+m^2 )
•  Ty=mx±a√(1+m^2 )
•  x=my±a√(1+m^2 )
•  None of these
Solution
y=mx±a√(1+m^2 )

Q9.The line 3x-2y=k meets the circle x^2+y^2+2gx+2fy+c=0 at only one point, if k^2=
•  20r^2
•  52r^2
•  52/9 r^2
•  20/9 r^2
Solution
52r^2

Q10. The line lx+my+n=0 will be a tangent to the circle x^2+y^2=a^2 if
•  n^2 (l^2+m^2)=a^2
•  a^2 (l^2+m^2)=n^2
•  n(l+m)=a
• a(l+m)=n
Solution
a^2 (l^2+m^2)=n^2 ## Want to know more

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Equation of Tangent, Condition for Tangency and the points of Contact - Basic Quiz
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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE
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