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The equation of a circle which is concentric to the given circle x2 + y2 - 4x - 6y - 3 = 0 and which touches the Y-axis is:
- a)x2 + y2 + 4x + 6y + 13 = 0
- b)x2 + y2 - 4x - 6y + 13 = 0
- c)x2 + y2 - 4x - 6y + 9 = 0
- d)x2 + y2 - 4x - 6y + 4 = 0
Correct answer is option 'C'. Can you explain this answer?
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Finding the equation of a circle concentric to given circle and tangent to x-axis
Given circle: x2 + y2 - 4x - 6y - 3 = 0
Step 1: Finding the center of the given circle
To find the center of the given circle, we need to complete the square for both x and y terms.
x2 - 4x + y2 - 6y = 3
(x - 2)2 - 4 + (y - 3)2 - 9 = 3
(x - 2)2 + (y - 3)2 = 16
The center of the given circle is (2, 3) and the radius is 4.
Step 2: Finding the equation of the required circle
Since the required circle is concentric to the given circle and tangent to x-axis, it must have the same center as the given circle.
Let the equation of the required circle be x2 + y2 + Dx + Ey + F = 0. Since it is tangent to x-axis, the distance from the center of the circle to the x-axis is equal to the radius of the circle.
Distance from center to x-axis = |E|/√(D2 + E2) = 4
Squaring both sides and simplifying, we get:
D2 + E2 = 16
Since the required circle is concentric to the given circle, the distance between their centers is zero.
Distance between centers = √[(2 + D/2)2 + (3 + E/2)2] = 0
Squaring both sides and simplifying, we get:
D2 + E2 + 8D + 12E + 13 = 0
Solving the above two equations simultaneously, we get:
D = -4 and E = -6
Therefore, the equation of the required circle is:
x2 + y2 - 4x - 6y + 4 = 0
Which can be simplified to:
x2 + y2 - 4x - 6y + 4 = (x - 2)2 + (y - 3)2 - 16 + 4
x2 + y2 - 4x - 6y + 4 = (x - 2)2 + (y - 3)2 - 12
Therefore, the equation of the required circle is:
x2 + y2 - 4x - 6y + 4 = 0
Option C is correct.
Given circle: x2 + y2 - 4x - 6y - 3 = 0
Step 1: Finding the center of the given circle
To find the center of the given circle, we need to complete the square for both x and y terms.
x2 - 4x + y2 - 6y = 3
(x - 2)2 - 4 + (y - 3)2 - 9 = 3
(x - 2)2 + (y - 3)2 = 16
The center of the given circle is (2, 3) and the radius is 4.
Step 2: Finding the equation of the required circle
Since the required circle is concentric to the given circle and tangent to x-axis, it must have the same center as the given circle.
Let the equation of the required circle be x2 + y2 + Dx + Ey + F = 0. Since it is tangent to x-axis, the distance from the center of the circle to the x-axis is equal to the radius of the circle.
Distance from center to x-axis = |E|/√(D2 + E2) = 4
Squaring both sides and simplifying, we get:
D2 + E2 = 16
Since the required circle is concentric to the given circle, the distance between their centers is zero.
Distance between centers = √[(2 + D/2)2 + (3 + E/2)2] = 0
Squaring both sides and simplifying, we get:
D2 + E2 + 8D + 12E + 13 = 0
Solving the above two equations simultaneously, we get:
D = -4 and E = -6
Therefore, the equation of the required circle is:
x2 + y2 - 4x - 6y + 4 = 0
Which can be simplified to:
x2 + y2 - 4x - 6y + 4 = (x - 2)2 + (y - 3)2 - 16 + 4
x2 + y2 - 4x - 6y + 4 = (x - 2)2 + (y - 3)2 - 12
Therefore, the equation of the required circle is:
x2 + y2 - 4x - 6y + 4 = 0
Option C is correct.
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The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 0Correct answer is option 'C'. Can you explain this answer?
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The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 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 The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 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 The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 0Correct answer is option 'C'. Can you explain this answer?.
The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 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 The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 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 The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 0Correct answer is option 'C'. Can you explain this answer?.
Solutions for The equation of a circle which is concentric to the given circlex2 + y2 - 4x - 6y - 3 = 0and which touches the Y-axis is:a)x2 + y2+4x +6y +13 = 0b)x2 + y2-4x -6y +13 = 0c)x2 + y2-4x -6y + 9= 0d)x2 + y2-4x -6y + 4 = 0Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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