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The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:
- a)x2 + y2 - 2ax - 2by = 0
- b)x2 + y2 - ax - by = a2 + b2
- c)x2 + y2 - ax - by = 0
- d)x2 + y2 + 4x + 6y + 13 = 0
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The equation of the circle passing through (0, 0) and making intercept...
Let the equation of the required circle be (x – h)2 + (y – k)2 = r2.
Since the centre of the circle passes through (0, 0),
(0 – h)2 + (0 – k)2 = r2
⇒ h2 + k2 = r2
The equation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)2 + (0 – k)2 = h2 + k2 …………… (1)
(0 – h)2 + (b – k)2 = h2 + k2 ………… (2)
From equation (1), we obtain a2 – 2ah + h2 + k2 = h2 + k2
⇒ a2 – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.
From equation (2), we obtain h2 + b2 – 2bk + k2 = h2 + k2
⇒ b2 – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or(b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =b/2. Thus, the equation of the required circle is
Since the centre of the circle passes through (0, 0),
(0 – h)2 + (0 – k)2 = r2
⇒ h2 + k2 = r2
The equation of the circle now becomes (x – h)2 + (y – k)2 = h2 + k2.
It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,
(a – h)2 + (0 – k)2 = h2 + k2 …………… (1)
(0 – h)2 + (b – k)2 = h2 + k2 ………… (2)
From equation (1), we obtain a2 – 2ah + h2 + k2 = h2 + k2
⇒ a2 – 2ah = 0
⇒ a(a – 2h) = 0
⇒ a = 0 or (a – 2h) = 0
However, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.
From equation (2), we obtain h2 + b2 – 2bk + k2 = h2 + k2
⇒ b2 – 2bk = 0
⇒ b(b – 2k) = 0
⇒ b = 0 or(b – 2k) = 0
However, b ≠ 0; hence, (b – 2k) = 0 ⇒ k =b/2. Thus, the equation of the required circle is
Community Answer
The equation of the circle passing through (0, 0) and making intercept...
Given information:
- The circle passes through the point (0, 0).
- The circle intercepts on the coordinate axes are a and b.
To find the equation of the circle, we can use the standard form of the equation of a circle:
(x - h)² + (y - k)² = r²
where (h, k) is the center of the circle and r is the radius.
Let's find the center of the circle first:
Since the circle passes through the point (0, 0), the center must lie on the perpendicular bisectors of the intercepts on the coordinate axes.
The perpendicular bisector of the x-axis intercept (a, 0) is the line y = 0.
The perpendicular bisector of the y-axis intercept (0, b) is the line x = 0.
Therefore, the center of the circle is the intersection of these two lines, which is the point (0, 0).
Now, let's find the radius of the circle:
The radius can be found using the distance formula between the center (0, 0) and any point on the circle. Let's take the x-axis intercept (a, 0).
The distance between (0, 0) and (a, 0) is given by:
r = √[(a - 0)² + (0 - 0)²]
= √[a²]
= a
So, the radius of the circle is a.
Therefore, the equation of the circle is:
(x - 0)² + (y - 0)² = a²
x² + y² = a²
However, none of the given options match this equation.
Let's check the options again:
a) x² + y² - 2ax - 2by = 0
b) x² + y² - ax - by = a² + b²
c) x² + y² - ax - by = 0
d) x² + y² + 4x + 6y + 13 = 0
Option C (x² + y² - ax - by = 0) is the closest match to the equation we derived (x² + y² = a²).
Therefore, the correct answer is option C.
- The circle passes through the point (0, 0).
- The circle intercepts on the coordinate axes are a and b.
To find the equation of the circle, we can use the standard form of the equation of a circle:
(x - h)² + (y - k)² = r²
where (h, k) is the center of the circle and r is the radius.
Let's find the center of the circle first:
Since the circle passes through the point (0, 0), the center must lie on the perpendicular bisectors of the intercepts on the coordinate axes.
The perpendicular bisector of the x-axis intercept (a, 0) is the line y = 0.
The perpendicular bisector of the y-axis intercept (0, b) is the line x = 0.
Therefore, the center of the circle is the intersection of these two lines, which is the point (0, 0).
Now, let's find the radius of the circle:
The radius can be found using the distance formula between the center (0, 0) and any point on the circle. Let's take the x-axis intercept (a, 0).
The distance between (0, 0) and (a, 0) is given by:
r = √[(a - 0)² + (0 - 0)²]
= √[a²]
= a
So, the radius of the circle is a.
Therefore, the equation of the circle is:
(x - 0)² + (y - 0)² = a²
x² + y² = a²
However, none of the given options match this equation.
Let's check the options again:
a) x² + y² - 2ax - 2by = 0
b) x² + y² - ax - by = a² + b²
c) x² + y² - ax - by = 0
d) x² + y² + 4x + 6y + 13 = 0
Option C (x² + y² - ax - by = 0) is the closest match to the equation we derived (x² + y² = a²).
Therefore, the correct answer is option C.
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The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:a)x2 + y2 - 2ax - 2by= 0b)x2 + y2 - ax - by= a2 + b2c)x2 + y2 - ax - by = 0d)x2 + y2+4x +6y + 13 = 0Correct answer is option 'C'. 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 passing through (0, 0) and making intercepts a and b on the coordinate axes is:a)x2 + y2 - 2ax - 2by= 0b)x2 + y2 - ax - by= a2 + b2c)x2 + y2 - ax - by = 0d)x2 + y2+4x +6y + 13 = 0Correct answer is option 'C'. 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 passing through (0, 0) and making intercepts a and b on the coordinate axes is:a)x2 + y2 - 2ax - 2by= 0b)x2 + y2 - ax - by= a2 + b2c)x2 + y2 - ax - by = 0d)x2 + y2+4x +6y + 13 = 0Correct answer is option 'C'. Can you explain this answer?.
The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:a)x2 + y2 - 2ax - 2by= 0b)x2 + y2 - ax - by= a2 + b2c)x2 + y2 - ax - by = 0d)x2 + y2+4x +6y + 13 = 0Correct answer is option 'C'. 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 passing through (0, 0) and making intercepts a and b on the coordinate axes is:a)x2 + y2 - 2ax - 2by= 0b)x2 + y2 - ax - by= a2 + b2c)x2 + y2 - ax - by = 0d)x2 + y2+4x +6y + 13 = 0Correct answer is option 'C'. 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 passing through (0, 0) and making intercepts a and b on the coordinate axes is:a)x2 + y2 - 2ax - 2by= 0b)x2 + y2 - ax - by= a2 + b2c)x2 + y2 - ax - by = 0d)x2 + y2+4x +6y + 13 = 0Correct answer is option 'C'. Can you explain this answer?.
Solutions for The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:a)x2 + y2 - 2ax - 2by= 0b)x2 + y2 - ax - by= a2 + b2c)x2 + y2 - ax - by = 0d)x2 + y2+4x +6y + 13 = 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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