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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?
Verified 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
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The equation of the circle passing through (0, 0) and making intercept...
The equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes is:
To find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes, we can use the general equation of a circle.
The general equation of a circle is given by:
(x - h)² + (y - k)² = r²
where (h, k) is the center of the circle and r is the radius.
Step 1: Find the center of the circle:
Since the circle passes through (0, 0), the center of the circle is the midpoint of the line segment joining (0, a) and (b, 0).
The midpoint formula is given by:
(h, k) = ((0 + b)/2, (a + 0)/2) = (b/2, a/2)
So, the center of the circle is (b/2, a/2).
Step 2: Find the radius of the circle:
The radius of the circle can be found using the distance formula between the center of the circle and any point on the circle. Let's use the point (b, 0) on the x-axis.
The distance formula is given by:
r = sqrt((x2 - x1)² + (y2 - y1)²)
r = sqrt((b - b/2)² + (0 - a/2)²)
r = sqrt((b/2)² + (a/2)²)
r = sqrt(b²/4 + a²/4)
r = sqrt((a² + b²)/4)
So, the radius of the circle is sqrt((a² + b²)/4).
Step 3: Write the equation of the circle:
Now that we have the center of the circle (b/2, a/2) and the radius sqrt((a² + b²)/4), we can substitute these values into the general equation of a circle to get the equation of the circle.
(x - b/2)² + (y - a/2)² = (sqrt((a² + b²)/4))²
(x - b/2)² + (y - a/2)² = (a² + b²)/4
Multiplying both sides of the equation by 4, we get:
4(x - b/2)² + 4(y - a/2)² = a² + b²
Expanding and simplifying, we get:
4x² - 4bx + 4y² - 4ay = a² + b²
Dividing both sides of the equation by 4, we get:
x² - bx + y² - ay = (a² + b²)/4
Multiplying both sides of the equation by -1, we get:
-bx - ay + x² + y² = -(a² + b²)/4
Rearranging the terms, we get:
x² + y² - bx - ay = -(a² + b²)/4
Multiplying both sides of the equation by -4, we get:
-4x² - 4y² + 4bx + 4ay
To find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes, we can use the general equation of a circle.
The general equation of a circle is given by:
(x - h)² + (y - k)² = r²
where (h, k) is the center of the circle and r is the radius.
Step 1: Find the center of the circle:
Since the circle passes through (0, 0), the center of the circle is the midpoint of the line segment joining (0, a) and (b, 0).
The midpoint formula is given by:
(h, k) = ((0 + b)/2, (a + 0)/2) = (b/2, a/2)
So, the center of the circle is (b/2, a/2).
Step 2: Find the radius of the circle:
The radius of the circle can be found using the distance formula between the center of the circle and any point on the circle. Let's use the point (b, 0) on the x-axis.
The distance formula is given by:
r = sqrt((x2 - x1)² + (y2 - y1)²)
r = sqrt((b - b/2)² + (0 - a/2)²)
r = sqrt((b/2)² + (a/2)²)
r = sqrt(b²/4 + a²/4)
r = sqrt((a² + b²)/4)
So, the radius of the circle is sqrt((a² + b²)/4).
Step 3: Write the equation of the circle:
Now that we have the center of the circle (b/2, a/2) and the radius sqrt((a² + b²)/4), we can substitute these values into the general equation of a circle to get the equation of the circle.
(x - b/2)² + (y - a/2)² = (sqrt((a² + b²)/4))²
(x - b/2)² + (y - a/2)² = (a² + b²)/4
Multiplying both sides of the equation by 4, we get:
4(x - b/2)² + 4(y - a/2)² = a² + b²
Expanding and simplifying, we get:
4x² - 4bx + 4y² - 4ay = a² + b²
Dividing both sides of the equation by 4, we get:
x² - bx + y² - ay = (a² + b²)/4
Multiplying both sides of the equation by -1, we get:
-bx - ay + x² + y² = -(a² + b²)/4
Rearranging the terms, we get:
x² + y² - bx - ay = -(a² + b²)/4
Multiplying both sides of the equation by -4, we get:
-4x² - 4y² + 4bx + 4ay
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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?
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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 2024 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 2024 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?.
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