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PASSAGE-3
A tangent PT is drawn to the circle x2 + y2 = 4 at the point
. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)
A tangent PT is drawn to the circle x2 + y2 = 4 at the point
. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L is
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point....
Equation of PT : S1 = 0
: √3 x + y = 4
Slope of PT = -√3
Slope of line L = 1/√3
Equation of line L : y = x/√3 + c
x - √3 y + √3c = 0
For this line to be a tangent to the given circle put the condition that the perpendicular distance from the centre (3,0) must be equal to the radius 1 and solve for c.
You will get c = -1/√3 (or) c = -5/√3
Substituting back in the line equation we will get the equation of L as,
x - √3y = 1 and
x - √3y = 5
: √3 x + y = 4
Slope of PT = -√3
Slope of line L = 1/√3
Equation of line L : y = x/√3 + c
x - √3 y + √3c = 0
For this line to be a tangent to the given circle put the condition that the perpendicular distance from the centre (3,0) must be equal to the radius 1 and solve for c.
You will get c = -1/√3 (or) c = -5/√3
Substituting back in the line equation we will get the equation of L as,
x - √3y = 1 and
x - √3y = 5
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PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point....
Equation of tangent PT to the circle x2 + y2 = 4 at the point 
Let the line L, perpendicular to tangent PT be
As it is tangent to the circle (x – 3)2 + y2 = 1
∴ length of perpendicular from centre of circle to the tangent = radius of circle.
= – 1 or – 5∴ Equation of L can be 
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PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?
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PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?.
PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?.
Solutions for PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer?, a detailed solution for PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of PASSAGE-3A tangent PT is drawn to the circle x2 + y2 = 4 at the point. A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. (2012)Q. A possible equation of L isa)b)c)d)Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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