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The locus of the mid-points of the chords of the circle x2 + y2 – 2x – 4y – 11 = 0 which subtend 60º at the centre is
- a)x2 + y2 – 4x – 2y – 7 = 0
- b)x2 + y2 + 4x + 2y – 7 = 0
- c)x2 + y2 – 2x – 4y – 7 = 0
- d)x2 + y2 + 2x + 4y + 7 = 0
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The locus of the mid-points of the chords of the circle x2+ y2–2...
Let AB be the chord of the circle and P be the midpoint of AB.
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a right-angled triangle.
Now AC=BC= radius.
The equation of the give circle can be written as
(x−1)2+(y−2)2=16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)½]/2
= 2(3)1/2 units
Therefore, PC=2(3)1/2
⇒ PC2=12
⇒ (x−1)2+(y−2)2=12
⇒ x2+y2−2x−4y+5=12
⇒ x2+y2−2x−4y−7=0
It is known that perpendicular from the center bisects a chord.
Thus △ACP is a right-angled triangle.
Now AC=BC= radius.
The equation of the give circle can be written as
(x−1)2+(y−2)2=16
Hence, centre C=(1,2) and radius =r=4 units.
PC=ACsin60degree
= rsin60degree
= 4([2(3)½]/2
= 2(3)1/2 units
Therefore, PC=2(3)1/2
⇒ PC2=12
⇒ (x−1)2+(y−2)2=12
⇒ x2+y2−2x−4y+5=12
⇒ x2+y2−2x−4y−7=0
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The locus of the mid-points of the chords of the circle x2+ y2–2...
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The locus of the mid-points of the chords of the circle x2+ y2–2...
The equation of the given circle is x^2 + y^2 = r^2, where r is the radius of the circle.
Let (h, k) be the coordinates of the mid-point of a chord of the circle.
Since the mid-point of a chord divides the chord into two equal parts, we can consider one of the parts as a line segment and find its equation.
Let (x1, y1) and (x2, y2) be the coordinates of the endpoints of the chord. The mid-point coordinates can be found using the midpoint formula:
(h, k) = ((x1 + x2)/2, (y1 + y2)/2)
Simplifying, we get:
2h = x1 + x2
2k = y1 + y2
Squaring both sides of the equations, we get:
(2h)^2 = (x1 + x2)^2
(2k)^2 = (y1 + y2)^2
Expanding, we get:
4h^2 = x1^2 + 2x1x2 + x2^2
4k^2 = y1^2 + 2y1y2 + y2^2
Adding both equations together, we get:
4h^2 + 4k^2 = x1^2 + 2x1x2 + x2^2 + y1^2 + 2y1y2 + y2^2
Since the points (x1, y1) and (x2, y2) lie on the circle, we can substitute x1^2 + y1^2 = r^2 and x2^2 + y2^2 = r^2:
4h^2 + 4k^2 = 2r^2 + 2x1x2 + 2y1y2
Dividing both sides by 4, we get:
h^2 + k^2 = (r^2 + x1x2 + y1y2)/2
Since r^2, x1x2, and y1y2 are constants, the equation h^2 + k^2 = constant represents the locus of mid-points of chords of the given circle.
Therefore, the locus of the mid-points of the chords of the circle x^2 + y^2 = r^2 is h^2 + k^2 = constant.
Let (h, k) be the coordinates of the mid-point of a chord of the circle.
Since the mid-point of a chord divides the chord into two equal parts, we can consider one of the parts as a line segment and find its equation.
Let (x1, y1) and (x2, y2) be the coordinates of the endpoints of the chord. The mid-point coordinates can be found using the midpoint formula:
(h, k) = ((x1 + x2)/2, (y1 + y2)/2)
Simplifying, we get:
2h = x1 + x2
2k = y1 + y2
Squaring both sides of the equations, we get:
(2h)^2 = (x1 + x2)^2
(2k)^2 = (y1 + y2)^2
Expanding, we get:
4h^2 = x1^2 + 2x1x2 + x2^2
4k^2 = y1^2 + 2y1y2 + y2^2
Adding both equations together, we get:
4h^2 + 4k^2 = x1^2 + 2x1x2 + x2^2 + y1^2 + 2y1y2 + y2^2
Since the points (x1, y1) and (x2, y2) lie on the circle, we can substitute x1^2 + y1^2 = r^2 and x2^2 + y2^2 = r^2:
4h^2 + 4k^2 = 2r^2 + 2x1x2 + 2y1y2
Dividing both sides by 4, we get:
h^2 + k^2 = (r^2 + x1x2 + y1y2)/2
Since r^2, x1x2, and y1y2 are constants, the equation h^2 + k^2 = constant represents the locus of mid-points of chords of the given circle.
Therefore, the locus of the mid-points of the chords of the circle x^2 + y^2 = r^2 is h^2 + k^2 = constant.
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The locus of the mid-points of the chords of the circle x2+ y2–2x –4y –11 = 0 which subtend 60º at the centre isa)x2+ y2–4x –2y –7 = 0b)x2+ y2+ 4x + 2y –7 = 0c)x2+ y2–2x –4y –7 = 0d)x2+ y2+ 2x + 4y + 7 = 0Correct answer is option 'C'. Can you explain this answer?
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The locus of the mid-points of the chords of the circle x2+ y2–2x –4y –11 = 0 which subtend 60º at the centre isa)x2+ y2–4x –2y –7 = 0b)x2+ y2+ 4x + 2y –7 = 0c)x2+ y2–2x –4y –7 = 0d)x2+ y2+ 2x + 4y + 7 = 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 locus of the mid-points of the chords of the circle x2+ y2–2x –4y –11 = 0 which subtend 60º at the centre isa)x2+ y2–4x –2y –7 = 0b)x2+ y2+ 4x + 2y –7 = 0c)x2+ y2–2x –4y –7 = 0d)x2+ y2+ 2x + 4y + 7 = 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 locus of the mid-points of the chords of the circle x2+ y2–2x –4y –11 = 0 which subtend 60º at the centre isa)x2+ y2–4x –2y –7 = 0b)x2+ y2+ 4x + 2y –7 = 0c)x2+ y2–2x –4y –7 = 0d)x2+ y2+ 2x + 4y + 7 = 0Correct answer is option 'C'. Can you explain this answer?.
The locus of the mid-points of the chords of the circle x2+ y2–2x –4y –11 = 0 which subtend 60º at the centre isa)x2+ y2–4x –2y –7 = 0b)x2+ y2+ 4x + 2y –7 = 0c)x2+ y2–2x –4y –7 = 0d)x2+ y2+ 2x + 4y + 7 = 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 locus of the mid-points of the chords of the circle x2+ y2–2x –4y –11 = 0 which subtend 60º at the centre isa)x2+ y2–4x –2y –7 = 0b)x2+ y2+ 4x + 2y –7 = 0c)x2+ y2–2x –4y –7 = 0d)x2+ y2+ 2x + 4y + 7 = 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 locus of the mid-points of the chords of the circle x2+ y2–2x –4y –11 = 0 which subtend 60º at the centre isa)x2+ y2–4x –2y –7 = 0b)x2+ y2+ 4x + 2y –7 = 0c)x2+ y2–2x –4y –7 = 0d)x2+ y2+ 2x + 4y + 7 = 0Correct answer is option 'C'. Can you explain this answer?.
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