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The equation of circle of radius 5 units touches the coordinates axes in the second quadrant is:
- a)x2 + y2 + 10x – 10y + 25 = 0
- b)x2 + y2 – 10x – 10y + 25 = 0
- c)x2 + y2 + 10x + 10y + 25 = 0
- d)x2 + y2 – 10x – 10y – 25 = 0
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The equation of circle of radius 5 units touches the coordinates axes ...
If the circle lies in second quadrant
The equation of a circle touches both the coordinate axes and has radius a is
x2 + y2 + 2ax - 2ay + a2 = 0
Radius of circle, a = 5
x2 + y2 + 10x - 10y + 25 = 0
The equation of a circle touches both the coordinate axes and has radius a is
x2 + y2 + 2ax - 2ay + a2 = 0
Radius of circle, a = 5
x2 + y2 + 10x - 10y + 25 = 0
Most Upvoted Answer
The equation of circle of radius 5 units touches the coordinates axes ...
General form for that one is :X²+Y²+2ax-2ay+a²=0
just substitute the values then u will get the required ans.
just substitute the values then u will get the required ans.
Community Answer
The equation of circle of radius 5 units touches the coordinates axes ...
To find the equation of a circle that touches the coordinate axes in the second quadrant, we need to find the center of the circle. Since the circle touches the x-axis and y-axis, the center must lie on both axes.
Let (h, k) be the center of the circle. Since the circle touches the x-axis, the distance from the center to the x-axis is equal to the radius (5 units). This means that the y-coordinate of the center is 5 units.
Since the circle touches the y-axis, the distance from the center to the y-axis is equal to the radius (5 units). This means that the x-coordinate of the center is -5 units.
So, the center of the circle is (-5, 5).
The equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2.
Substituting the values, we have (x - (-5))^2 + (y - 5)^2 = 5^2.
Simplifying, we get (x + 5)^2 + (y - 5)^2 = 25.
So, the equation of the circle is (x + 5)^2 + (y - 5)^2 = 25.
Let (h, k) be the center of the circle. Since the circle touches the x-axis, the distance from the center to the x-axis is equal to the radius (5 units). This means that the y-coordinate of the center is 5 units.
Since the circle touches the y-axis, the distance from the center to the y-axis is equal to the radius (5 units). This means that the x-coordinate of the center is -5 units.
So, the center of the circle is (-5, 5).
The equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2.
Substituting the values, we have (x - (-5))^2 + (y - 5)^2 = 5^2.
Simplifying, we get (x + 5)^2 + (y - 5)^2 = 25.
So, the equation of the circle is (x + 5)^2 + (y - 5)^2 = 25.
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The equation of circle of radius 5 units touches the coordinates axes in the second quadrant is:a)x2+ y2+ 10x – 10y + 25 = 0b)x2+ y2– 10x – 10y + 25 = 0c)x2+ y2+ 10x + 10y + 25 = 0d)x2+ y2– 10x – 10y –25 = 0Correct answer is option 'A'. Can you explain this answer?
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The equation of circle of radius 5 units touches the coordinates axes in the second quadrant is:a)x2+ y2+ 10x – 10y + 25 = 0b)x2+ y2– 10x – 10y + 25 = 0c)x2+ y2+ 10x + 10y + 25 = 0d)x2+ y2– 10x – 10y –25 = 0Correct answer is option 'A'. 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 circle of radius 5 units touches the coordinates axes in the second quadrant is:a)x2+ y2+ 10x – 10y + 25 = 0b)x2+ y2– 10x – 10y + 25 = 0c)x2+ y2+ 10x + 10y + 25 = 0d)x2+ y2– 10x – 10y –25 = 0Correct answer is option 'A'. 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 circle of radius 5 units touches the coordinates axes in the second quadrant is:a)x2+ y2+ 10x – 10y + 25 = 0b)x2+ y2– 10x – 10y + 25 = 0c)x2+ y2+ 10x + 10y + 25 = 0d)x2+ y2– 10x – 10y –25 = 0Correct answer is option 'A'. Can you explain this answer?.
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