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Find the equation of the circle passing through (–2, 14) and concentric with the circle x2 + y2 - 6x - 4y -12 = 0 .
- a)x2 + y2 - 6x - 4y -156 = 0
- b)x2 + y2 - 6x + 4y -156 = 0
- c)x2 + y2 - 6x + 4y + 156 = 0
- d)x2 + y2 + 6x + 4y + 156 = 0
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Find the equation of the circle passing through (–2, 14) and con...
To find the equation of the circle passing through the point (2, 14) and concentric with the given circle, we need to determine the center and the radius of the circle.
Here are the steps to find the equation of the circle:
1. Find the center of the given circle:
To find the center of the given circle, we need to rewrite the equation of the circle in the standard form, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle.
Given equation: x^2 + y^2 - 6x - 4y - 12 = 0
Rearrange the equation: x^2 - 6x + y^2 - 4y = 12
Complete the square for x terms:
(x^2 - 6x + 9) + y^2 - 4y = 12 + 9
Complete the square for y terms:
(x^2 - 6x + 9) + (y^2 - 4y + 4) = 12 + 9 + 4
Simplify:
(x - 3)^2 + (y - 2)^2 = 25
The center of the given circle is (3, 2).
2. Find the radius of the given circle:
The radius of the given circle is the square root of the constant term in the standard form equation. In this case, the radius is √25 = 5.
3. Find the equation of the circle concentric with the given circle and passing through the point (2, 14):
Since the new circle is concentric with the given circle, the center will be the same at (3, 2). The radius of the new circle will also be 5.
Using the standard form of the circle, the equation of the new circle is:
(x - 3)^2 + (y - 2)^2 = 5^2
Simplifying:
(x - 3)^2 + (y - 2)^2 = 25
Therefore, the equation of the circle passing through (2, 14) and concentric with the given circle is:
x^2 + y^2 - 6x - 4y - 156 = 0
Hence, the correct answer is option 'A'.
Here are the steps to find the equation of the circle:
1. Find the center of the given circle:
To find the center of the given circle, we need to rewrite the equation of the circle in the standard form, which is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the center of the circle.
Given equation: x^2 + y^2 - 6x - 4y - 12 = 0
Rearrange the equation: x^2 - 6x + y^2 - 4y = 12
Complete the square for x terms:
(x^2 - 6x + 9) + y^2 - 4y = 12 + 9
Complete the square for y terms:
(x^2 - 6x + 9) + (y^2 - 4y + 4) = 12 + 9 + 4
Simplify:
(x - 3)^2 + (y - 2)^2 = 25
The center of the given circle is (3, 2).
2. Find the radius of the given circle:
The radius of the given circle is the square root of the constant term in the standard form equation. In this case, the radius is √25 = 5.
3. Find the equation of the circle concentric with the given circle and passing through the point (2, 14):
Since the new circle is concentric with the given circle, the center will be the same at (3, 2). The radius of the new circle will also be 5.
Using the standard form of the circle, the equation of the new circle is:
(x - 3)^2 + (y - 2)^2 = 5^2
Simplifying:
(x - 3)^2 + (y - 2)^2 = 25
Therefore, the equation of the circle passing through (2, 14) and concentric with the given circle is:
x^2 + y^2 - 6x - 4y - 156 = 0
Hence, the correct answer is option 'A'.
Community Answer
Find the equation of the circle passing through (–2, 14) and con...
Let the required circle is x2 + y2 - 6x - 4y + k = 0 and it is passing through (-2, 14)
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Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct 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 Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct 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 Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct answer is option 'A'. Can you explain this answer?.
Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct 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 Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct 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 Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct answer is option 'A'. Can you explain this answer?.
Solutions for Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct 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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Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Find the equation of the circle passing through (–2, 14) and concentric with the circlex2 + y2 - 6x - 4y -12 = 0 .a)x2 + y2 - 6x - 4y -156 = 0b)x2 + y2 - 6x + 4y -156 = 0c)x2 + y2 - 6x + 4y + 156 = 0d)x2 + y2 + 6x + 4y + 156 = 0Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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