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A pair of tangents are drawn from the origin to the circle x2 + y2 + 20(x + y) + 20 = 0. The equation of the pair of tangents is
- a)x2 + y2 + 5xy = 0
- b)x2 + y2 + 10xy = 0
- c)2x2 + 2y2 + 5xy = 0
- d)2x2 + 2y2– 5xy = 0
Correct answer is option 'C'. Can you explain this answer?
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A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(...
Given equation of circle is
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A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(...
To find the equation of the pair of tangents, we can use the fact that the tangents from the origin to a circle are perpendicular to the radius at the point of tangency.
First, let's rewrite the equation of the circle in standard form:
x^2 + y^2 + 20x + 20y - 20 = 0
To find the coordinates of the point of tangency, we need to find the radius of the circle. The radius is the perpendicular distance from the origin to the center of the circle. We can find the center by completing the square for both x and y terms:
x^2 + 20x + y^2 + 20y - 20 = 0
(x^2 + 20x + 100) + (y^2 + 20y + 100) - 20 - 100 - 100 = 0
(x + 10)^2 + (y + 10)^2 = 240
Since the radius is the square root of 240, which is 4√15, the coordinates of the point of tangency are (-4√15, -4√15).
The slope of the radius at the point of tangency is perpendicular to the tangent line. The slope of the radius can be found using the formula:
m = (y2 - y1) / (x2 - x1)
where (x1, y1) = (0, 0) (origin) and (x2, y2) = (-4√15, -4√15) (point of tangency).
m = (-4√15 - 0) / (-4√15 - 0)
m = 4√15 / 4√15
m = 1
Now we can use the point-slope form of the equation of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
y - (-4√15) = 1(x - (-4√15))
y + 4√15 = x + 4√15
y = x + 8√15
Since the tangent line passes through the origin (0, 0), the equation can be simplified to:
y = x
Therefore, the equation of the pair of tangents is:
Answer: d) 2x^2 - 2y^2 = 0
First, let's rewrite the equation of the circle in standard form:
x^2 + y^2 + 20x + 20y - 20 = 0
To find the coordinates of the point of tangency, we need to find the radius of the circle. The radius is the perpendicular distance from the origin to the center of the circle. We can find the center by completing the square for both x and y terms:
x^2 + 20x + y^2 + 20y - 20 = 0
(x^2 + 20x + 100) + (y^2 + 20y + 100) - 20 - 100 - 100 = 0
(x + 10)^2 + (y + 10)^2 = 240
Since the radius is the square root of 240, which is 4√15, the coordinates of the point of tangency are (-4√15, -4√15).
The slope of the radius at the point of tangency is perpendicular to the tangent line. The slope of the radius can be found using the formula:
m = (y2 - y1) / (x2 - x1)
where (x1, y1) = (0, 0) (origin) and (x2, y2) = (-4√15, -4√15) (point of tangency).
m = (-4√15 - 0) / (-4√15 - 0)
m = 4√15 / 4√15
m = 1
Now we can use the point-slope form of the equation of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
y - (-4√15) = 1(x - (-4√15))
y + 4√15 = x + 4√15
y = x + 8√15
Since the tangent line passes through the origin (0, 0), the equation can be simplified to:
y = x
Therefore, the equation of the pair of tangents is:
Answer: d) 2x^2 - 2y^2 = 0
Community Answer
A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(...
(T) ^2=(S1)S
S(x, y) is eq of circle & S1=S(0,0)
S(x, y) is eq of circle & S1=S(0,0)
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A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(x + y) + 20 = 0. The equation of the pair of tangents isa)x2+ y2+ 5xy = 0b)x2+ y2+ 10xy = 0c)2x2+ 2y2+ 5xy = 0d)2x2+ 2y2–5xy = 0Correct answer is option 'C'. Can you explain this answer?
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A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(x + y) + 20 = 0. The equation of the pair of tangents isa)x2+ y2+ 5xy = 0b)x2+ y2+ 10xy = 0c)2x2+ 2y2+ 5xy = 0d)2x2+ 2y2–5xy = 0Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(x + y) + 20 = 0. The equation of the pair of tangents isa)x2+ y2+ 5xy = 0b)x2+ y2+ 10xy = 0c)2x2+ 2y2+ 5xy = 0d)2x2+ 2y2–5xy = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(x + y) + 20 = 0. The equation of the pair of tangents isa)x2+ y2+ 5xy = 0b)x2+ y2+ 10xy = 0c)2x2+ 2y2+ 5xy = 0d)2x2+ 2y2–5xy = 0Correct answer is option 'C'. Can you explain this answer?.
A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(x + y) + 20 = 0. The equation of the pair of tangents isa)x2+ y2+ 5xy = 0b)x2+ y2+ 10xy = 0c)2x2+ 2y2+ 5xy = 0d)2x2+ 2y2–5xy = 0Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(x + y) + 20 = 0. The equation of the pair of tangents isa)x2+ y2+ 5xy = 0b)x2+ y2+ 10xy = 0c)2x2+ 2y2+ 5xy = 0d)2x2+ 2y2–5xy = 0Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pair of tangents are drawn from the origin to the circle x2+ y2+ 20(x + y) + 20 = 0. The equation of the pair of tangents isa)x2+ y2+ 5xy = 0b)x2+ y2+ 10xy = 0c)2x2+ 2y2+ 5xy = 0d)2x2+ 2y2–5xy = 0Correct answer is option 'C'. Can you explain this answer?.
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