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A pair of tangents are drawn from the origin to the circle x2 + y2+ 20 (x + y) + 20 = 0, then the equation of the pair of tangent are
- a)x2 + y2 - 5xy = 0
- b)x2 + y2 + 2x + y = 0
- c)x2 + y2 – xy + 7 = 0
- d)2x2 + 2y2+ 5xy = 0
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A pair of tangents are drawn from the origin to the circle x2 + y2+ 20...
Equation of pair of tangents is given by SS1 = T2,
or S = x2 + y2 + 20 (x + y) + 20, S, = 20,
T = 10 (x + y) + 20 = 0
∴ SS1 = T2
⇒ 20 (x2 + y2 + 20 (x + y) + 20)
= 102 (x + y + 2)2
⇒ 4x2 + 4y2 + 10xy = 0
⇒ 2x2 + 2y2 + 5xy = 0
or S = x2 + y2 + 20 (x + y) + 20, S, = 20,
T = 10 (x + y) + 20 = 0
∴ SS1 = T2
⇒ 20 (x2 + y2 + 20 (x + y) + 20)
= 102 (x + y + 2)2
⇒ 4x2 + 4y2 + 10xy = 0
⇒ 2x2 + 2y2 + 5xy = 0
Most Upvoted Answer
A pair of tangents are drawn from the origin to the circle x2 + y2+ 20...
The equation of the given circle is x^2 + y^2 - 20x - 20y + 20 = 0.
To find the equation of the pair of tangents from the origin, we can use the fact that the tangents from a point outside the circle are equal in length.
Let the point of contact of the tangents on the circle be (a, b).
Since the tangents are drawn from the origin (0, 0), we can form a right triangle with the origin, point of contact (a, b), and the center of the circle (10, 10).
Using the distance formula, the length of the hypotenuse (OB) is equal to the radius of the circle, which is √(10^2 + 10^2) = √200 = 10√2.
The length of the other two sides of the triangle are OA = √(a^2 + b^2) and AB = √((a - 10)^2 + (b - 10)^2).
Since OA = AB, we have √(a^2 + b^2) = √((a - 10)^2 + (b - 10)^2).
Squaring both sides, we get a^2 + b^2 = (a - 10)^2 + (b - 10)^2.
Expanding and simplifying, we get a^2 + b^2 = a^2 - 20a + 100 + b^2 - 20b + 100.
Simplifying further, we get 20a + 20b = 200.
Dividing both sides by 20, we get a + b = 10.
This equation represents a line passing through the point (10, 0) and (0, 10).
The equation of this line can be written as x + y - 10 = 0.
Therefore, the equation of the pair of tangent lines is x + y - 10 = 0.
Thus, the correct answer is option b) x^2 + y^2 + 2x + y = 0.
To find the equation of the pair of tangents from the origin, we can use the fact that the tangents from a point outside the circle are equal in length.
Let the point of contact of the tangents on the circle be (a, b).
Since the tangents are drawn from the origin (0, 0), we can form a right triangle with the origin, point of contact (a, b), and the center of the circle (10, 10).
Using the distance formula, the length of the hypotenuse (OB) is equal to the radius of the circle, which is √(10^2 + 10^2) = √200 = 10√2.
The length of the other two sides of the triangle are OA = √(a^2 + b^2) and AB = √((a - 10)^2 + (b - 10)^2).
Since OA = AB, we have √(a^2 + b^2) = √((a - 10)^2 + (b - 10)^2).
Squaring both sides, we get a^2 + b^2 = (a - 10)^2 + (b - 10)^2.
Expanding and simplifying, we get a^2 + b^2 = a^2 - 20a + 100 + b^2 - 20b + 100.
Simplifying further, we get 20a + 20b = 200.
Dividing both sides by 20, we get a + b = 10.
This equation represents a line passing through the point (10, 0) and (0, 10).
The equation of this line can be written as x + y - 10 = 0.
Therefore, the equation of the pair of tangent lines is x + y - 10 = 0.
Thus, the correct answer is option b) x^2 + y^2 + 2x + y = 0.
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A pair of tangents are drawn from the origin to the circle x2 + y2+ 20 (x + y) + 20 = 0, then the equation of the pair of tangent area)x2 + y2 - 5xy = 0b)x2 + y2 + 2x + y = 0c)x2 + y2 – xy + 7 = 0d)2x2 + 2y2+ 5xy = 0Correct answer is option 'D'. 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, then the equation of the pair of tangent area)x2 + y2 - 5xy = 0b)x2 + y2 + 2x + y = 0c)x2 + y2 – xy + 7 = 0d)2x2 + 2y2+ 5xy = 0Correct answer is option 'D'. 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, then the equation of the pair of tangent area)x2 + y2 - 5xy = 0b)x2 + y2 + 2x + y = 0c)x2 + y2 – xy + 7 = 0d)2x2 + 2y2+ 5xy = 0Correct answer is option 'D'. 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, then the equation of the pair of tangent area)x2 + y2 - 5xy = 0b)x2 + y2 + 2x + y = 0c)x2 + y2 – xy + 7 = 0d)2x2 + 2y2+ 5xy = 0Correct answer is option 'D'. Can you explain this answer?.
A pair of tangents are drawn from the origin to the circle x2 + y2+ 20 (x + y) + 20 = 0, then the equation of the pair of tangent area)x2 + y2 - 5xy = 0b)x2 + y2 + 2x + y = 0c)x2 + y2 – xy + 7 = 0d)2x2 + 2y2+ 5xy = 0Correct answer is option 'D'. 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, then the equation of the pair of tangent area)x2 + y2 - 5xy = 0b)x2 + y2 + 2x + y = 0c)x2 + y2 – xy + 7 = 0d)2x2 + 2y2+ 5xy = 0Correct answer is option 'D'. 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, then the equation of the pair of tangent area)x2 + y2 - 5xy = 0b)x2 + y2 + 2x + y = 0c)x2 + y2 – xy + 7 = 0d)2x2 + 2y2+ 5xy = 0Correct answer is option 'D'. Can you explain this answer?.
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