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A circle always passes through the fixed points (a, 0) and (–a, 0).
Q.
If given points are the ends of diameter, then the equation of circle is
- a)x2 + y2 = a2
- b)x2 + y2 + a2 = 0
- c)x2 + y2 + 2x + 2y = a2
- d)x2 + y2 – 2x – 2y = a2
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A circle always passes through the fixed points (a, 0) and (a, 0).Q.If...
Explanation:
Given Information:
- The circle passes through the fixed points (a, 0) and (-a, 0).
- The given points are the ends of a diameter.
Equation of Circle:
- The equation of a circle with center (h, k) and radius r is given by: (x-h)^2 + (y-k)^2 = r^2.
Center of the Circle:
- The center of the circle will be the midpoint of the diameter, which is the average of the given points.
- Center = ((a + (-a))/2, (0 + 0)/2) = (0, 0).
Radius of the Circle:
- The distance between the center and any of the given points is the radius of the circle.
- Radius = √((0-a)^2 + (0-0)^2) = √(a^2) = a.
Equation of the Circle:
- Substituting the center (0, 0) and radius a into the general equation of a circle:
- (x-0)^2 + (y-0)^2 = a^2.
- Simplifying, we get: x^2 + y^2 = a^2.
Therefore, the correct equation of the circle passing through the fixed points (a, 0) and (-a, 0) is x^2 + y^2 = a^2 (Option A).
Given Information:
- The circle passes through the fixed points (a, 0) and (-a, 0).
- The given points are the ends of a diameter.
Equation of Circle:
- The equation of a circle with center (h, k) and radius r is given by: (x-h)^2 + (y-k)^2 = r^2.
Center of the Circle:
- The center of the circle will be the midpoint of the diameter, which is the average of the given points.
- Center = ((a + (-a))/2, (0 + 0)/2) = (0, 0).
Radius of the Circle:
- The distance between the center and any of the given points is the radius of the circle.
- Radius = √((0-a)^2 + (0-0)^2) = √(a^2) = a.
Equation of the Circle:
- Substituting the center (0, 0) and radius a into the general equation of a circle:
- (x-0)^2 + (y-0)^2 = a^2.
- Simplifying, we get: x^2 + y^2 = a^2.
Therefore, the correct equation of the circle passing through the fixed points (a, 0) and (-a, 0) is x^2 + y^2 = a^2 (Option A).
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Community Answer
A circle always passes through the fixed points (a, 0) and (a, 0).Q.If...
Explanation:
To find the equation of a circle passing through the fixed points (a, 0) and (-a, 0), we can use the general equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) is the center of the circle and r is the radius.
Step 1: Find the center of the circle.
- The x-coordinate of the center is the average of the x-coordinates of the two fixed points: (a + (-a))/2 = 0.
- The y-coordinate of the center is the average of the y-coordinates of the two fixed points: (0 + 0)/2 = 0.
- Therefore, the center of the circle is (0, 0).
Step 2: Find the radius of the circle.
- The radius is the distance from the center to any of the fixed points.
- The distance between (a, 0) and the center (0, 0) is given by the formula: √((a - 0)^2 + (0 - 0)^2) = √(a^2) = a.
- Therefore, the radius of the circle is a.
Step 3: Write the equation of the circle.
- Substituting the values of the center (h, k) = (0, 0) and the radius r = a into the general equation of a circle, we get:
(x - 0)^2 + (y - 0)^2 = a^2
x^2 + y^2 = a^2
Therefore, the equation of the circle passing through the fixed points (a, 0) and (-a, 0) is x^2 + y^2 = a^2.
Hence, the correct answer is A: x^2 + y^2 = a^2.
To find the equation of a circle passing through the fixed points (a, 0) and (-a, 0), we can use the general equation of a circle:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) is the center of the circle and r is the radius.
Step 1: Find the center of the circle.
- The x-coordinate of the center is the average of the x-coordinates of the two fixed points: (a + (-a))/2 = 0.
- The y-coordinate of the center is the average of the y-coordinates of the two fixed points: (0 + 0)/2 = 0.
- Therefore, the center of the circle is (0, 0).
Step 2: Find the radius of the circle.
- The radius is the distance from the center to any of the fixed points.
- The distance between (a, 0) and the center (0, 0) is given by the formula: √((a - 0)^2 + (0 - 0)^2) = √(a^2) = a.
- Therefore, the radius of the circle is a.
Step 3: Write the equation of the circle.
- Substituting the values of the center (h, k) = (0, 0) and the radius r = a into the general equation of a circle, we get:
(x - 0)^2 + (y - 0)^2 = a^2
x^2 + y^2 = a^2
Therefore, the equation of the circle passing through the fixed points (a, 0) and (-a, 0) is x^2 + y^2 = a^2.
Hence, the correct answer is A: x^2 + y^2 = a^2.
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A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer?
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A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer?.
A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer? for Defence 2024 is part of Defence preparation. The Question and answers have been prepared according to the Defence exam syllabus. Information about A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Defence 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer?.
Solutions for A circle always passes through the fixed points (a, 0) and (a, 0).Q.If given points are the ends of diameter, then the equation of circle isa)x2 + y2 = a2b)x2 + y2 + a2 = 0c)x2 + y2 + 2x + 2y = a2d)x2 + y2 2x 2y = a2Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Defence.
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