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The equation of the circle which touches the axis of y at the origin and passes through (3, 4) is
- a)4(x2 + y2) – 25x = 0
- b)3(x2 + y2) – 25x = 0
- c)2(x2 + y2) – 3x = 0
- d)4(x2 + y2) – 25x + 10 = 0
Correct answer is option 'B'. Can you explain this answer?
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The equation of a circle with center (h, k) and radius r is given by:
(x - h)^2 + (y - k)^2 = r^2
In this case, the circle touches the y-axis at the origin, which means the center of the circle lies on the x-axis. Therefore, the y-coordinate of the center is 0.
Let's call the x-coordinate of the center of the circle as h.
Plugging in the coordinates of the point (3, 4) into the equation, we get:
(3 - h)^2 + (4 - 0)^2 = r^2
Expanding and simplifying, we get:
9 - 6h + h^2 + 16 = r^2
Combining like terms, we have:
h^2 - 6h + 25 = r^2
So, the equation of the circle is:
(x - h)^2 + y^2 = h^2 - 6h + 25
Depending on the value of h, the equation could be different.
(x - h)^2 + (y - k)^2 = r^2
In this case, the circle touches the y-axis at the origin, which means the center of the circle lies on the x-axis. Therefore, the y-coordinate of the center is 0.
Let's call the x-coordinate of the center of the circle as h.
Plugging in the coordinates of the point (3, 4) into the equation, we get:
(3 - h)^2 + (4 - 0)^2 = r^2
Expanding and simplifying, we get:
9 - 6h + h^2 + 16 = r^2
Combining like terms, we have:
h^2 - 6h + 25 = r^2
So, the equation of the circle is:
(x - h)^2 + y^2 = h^2 - 6h + 25
Depending on the value of h, the equation could be different.
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The equation of the circle which touches the axis of y at the origin and passes through (3, 4) isa)4(x2+ y2) –25x = 0b)3(x2+ y2) –25x = 0c)2(x2+ y2) –3x = 0d)4(x2+ y2) –25x + 10 = 0Correct answer is option 'B'. 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 The equation of the circle which touches the axis of y at the origin and passes through (3, 4) isa)4(x2+ y2) –25x = 0b)3(x2+ y2) –25x = 0c)2(x2+ y2) –3x = 0d)4(x2+ y2) –25x + 10 = 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the circle which touches the axis of y at the origin and passes through (3, 4) isa)4(x2+ y2) –25x = 0b)3(x2+ y2) –25x = 0c)2(x2+ y2) –3x = 0d)4(x2+ y2) –25x + 10 = 0Correct answer is option 'B'. Can you explain this answer?.
The equation of the circle which touches the axis of y at the origin and passes through (3, 4) isa)4(x2+ y2) –25x = 0b)3(x2+ y2) –25x = 0c)2(x2+ y2) –3x = 0d)4(x2+ y2) –25x + 10 = 0Correct answer is option 'B'. 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 The equation of the circle which touches the axis of y at the origin and passes through (3, 4) isa)4(x2+ y2) –25x = 0b)3(x2+ y2) –25x = 0c)2(x2+ y2) –3x = 0d)4(x2+ y2) –25x + 10 = 0Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equation of the circle which touches the axis of y at the origin and passes through (3, 4) isa)4(x2+ y2) –25x = 0b)3(x2+ y2) –25x = 0c)2(x2+ y2) –3x = 0d)4(x2+ y2) –25x + 10 = 0Correct answer is option 'B'. Can you explain this answer?.
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