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Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2√2 and 2√5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to:
- a)√11
- b)√7
- c)√6
- d)√10
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Let the lengths of intercepts on x-axis and y-axis made by the circle ...
x2 + y2 + ax + 2ay + c = 0
⇒ a2 - c = 5 ...(2)
From (1) & (2),
∴ c = -1
Circle: x2 + y2 - 2x - 4y - 1 = 0
⇒ (x - 1)2 + (y - 2)2 = 6
mtangent = 2
Equation of tangent:
2x – y + = 0
∴ p = r
⇒ a2 - c = 5 ...(2)
From (1) & (2),
∴ c = -1
Circle: x2 + y2 - 2x - 4y - 1 = 0
⇒ (x - 1)2 + (y - 2)2 = 6
mtangent = 2
Equation of tangent:
2x – y + = 0
∴ p = r
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Community Answer
Let the lengths of intercepts on x-axis and y-axis made by the circle ...
Let's assume that the equation of the circle is x^2 + y^2 + ax + 2ay + c = 0, where a and c are constants.
To find the lengths of intercepts on the x-axis and y-axis, we need to find the x and y coordinates of the points where the circle intersects these axes.
Intercept on x-axis:
To find the x-coordinate of the point where the circle intersects the x-axis, we set y = 0 in the equation of the circle:
x^2 + ax + c = 0
Using the quadratic formula, we can solve for x:
x = (-a ± √(a^2 - 4c)) / 2
The length of the intercept on the x-axis is the distance between these two x-coordinates:
Length of intercept on x-axis = |x1 - x2| = |((-a + √(a^2 - 4c)) / 2) - ((-a - √(a^2 - 4c)) / 2)| = |√(a^2 - 4c)|
Intercept on y-axis:
To find the y-coordinate of the point where the circle intersects the y-axis, we set x = 0 in the equation of the circle:
y^2 + 2ay + c = 0
Using the quadratic formula, we can solve for y:
y = (-2a ± √(4a^2 - 4c)) / 2 = -a ± √(a^2 - c)
The length of the intercept on the y-axis is the distance between these two y-coordinates:
Length of intercept on y-axis = |y1 - y2| = |-a + √(a^2 - c) - (-a - √(a^2 - c))| = 2√(a^2 - c)
Therefore, the lengths of the intercepts on the x-axis and y-axis made by the circle x^2 + y^2 + ax + 2ay + c = 0 are |√(a^2 - 4c)| and 2√(a^2 - c), respectively.
To find the lengths of intercepts on the x-axis and y-axis, we need to find the x and y coordinates of the points where the circle intersects these axes.
Intercept on x-axis:
To find the x-coordinate of the point where the circle intersects the x-axis, we set y = 0 in the equation of the circle:
x^2 + ax + c = 0
Using the quadratic formula, we can solve for x:
x = (-a ± √(a^2 - 4c)) / 2
The length of the intercept on the x-axis is the distance between these two x-coordinates:
Length of intercept on x-axis = |x1 - x2| = |((-a + √(a^2 - 4c)) / 2) - ((-a - √(a^2 - 4c)) / 2)| = |√(a^2 - 4c)|
Intercept on y-axis:
To find the y-coordinate of the point where the circle intersects the y-axis, we set x = 0 in the equation of the circle:
y^2 + 2ay + c = 0
Using the quadratic formula, we can solve for y:
y = (-2a ± √(4a^2 - 4c)) / 2 = -a ± √(a^2 - c)
The length of the intercept on the y-axis is the distance between these two y-coordinates:
Length of intercept on y-axis = |y1 - y2| = |-a + √(a^2 - c) - (-a - √(a^2 - c))| = 2√(a^2 - c)
Therefore, the lengths of the intercepts on the x-axis and y-axis made by the circle x^2 + y^2 + ax + 2ay + c = 0 are |√(a^2 - 4c)| and 2√(a^2 - c), respectively.
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Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2√2 and 2√5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to:a)√11b)√7c)√6d)√10Correct answer is option 'C'. Can you explain this answer?
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Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2√2 and 2√5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to:a)√11b)√7c)√6d)√10Correct 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 Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2√2 and 2√5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to:a)√11b)√7c)√6d)√10Correct 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 Let the lengths of intercepts on x-axis and y-axis made by the circle x2 + y2 + ax + 2ay + c = 0, (a < 0) be 2√2 and 2√5, respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to:a)√11b)√7c)√6d)√10Correct answer is option 'C'. Can you explain this answer?.
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