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The locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 passing through the origin is
- a)x2+y2+x-3y=0
- b)x2+y2-x-3y=0
- c)x2+y2-x+3y=0
- d)x2+y2+x+3y=0
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 pa...
Solution:
Finding the equation of circle:
The given equation can be written as:
x2-2x+y2-6y=10
Completing the square, we get:
(x-1)2+(y-3)2=20
Hence, the given equation represents a circle with center (1,3) and radius √20.
Finding the equation of chords:
The chord passing through the origin has equation y=mx. Substituting this in the equation of circle, we get:
x2+(mx)2-2x-6mx-10=0
Simplifying, we get:
(x2+m2x2-2mx)+(-2x-6mx-10)=0
The x-coordinates of the points of intersection of this chord with the circle are the roots of this quadratic equation. By Vieta’s formulas, the sum of the roots is:
x1+x2=2m/m2+1
The midpoint of this chord has coordinates ((x1+x2)/2, (y1+y2)/2), which can be expressed in terms of m as:
((2m/m2+1)/2, m(2m/m2+1)/2)
Simplifying, we get:
(x,y)=(m2-1/2m, m/2(m2+1))
Eliminating m:
The equation of the locus of the midpoints can be obtained by eliminating m from the equations for x and y. We have:
x=(m2-1)/2m
y=m/2(m2+1)
Eliminating m, we get:
x2+y2-xy-x-y=0
Simplifying, we get:
(x-y)2-(x+y)=0
Substituting z=x-y, we get:
z2-z=0
Solving for z, we get z=0 or z=1.
Substituting back for z, we get two lines:
x-y=0 and x+y-1=0.
Hence, the locus of midpoints is given by the equation:
x2+y2-x-3y=0
Therefore, option 'B' is correct.
Finding the equation of circle:
The given equation can be written as:
x2-2x+y2-6y=10
Completing the square, we get:
(x-1)2+(y-3)2=20
Hence, the given equation represents a circle with center (1,3) and radius √20.
Finding the equation of chords:
The chord passing through the origin has equation y=mx. Substituting this in the equation of circle, we get:
x2+(mx)2-2x-6mx-10=0
Simplifying, we get:
(x2+m2x2-2mx)+(-2x-6mx-10)=0
The x-coordinates of the points of intersection of this chord with the circle are the roots of this quadratic equation. By Vieta’s formulas, the sum of the roots is:
x1+x2=2m/m2+1
The midpoint of this chord has coordinates ((x1+x2)/2, (y1+y2)/2), which can be expressed in terms of m as:
((2m/m2+1)/2, m(2m/m2+1)/2)
Simplifying, we get:
(x,y)=(m2-1/2m, m/2(m2+1))
Eliminating m:
The equation of the locus of the midpoints can be obtained by eliminating m from the equations for x and y. We have:
x=(m2-1)/2m
y=m/2(m2+1)
Eliminating m, we get:
x2+y2-xy-x-y=0
Simplifying, we get:
(x-y)2-(x+y)=0
Substituting z=x-y, we get:
z2-z=0
Solving for z, we get z=0 or z=1.
Substituting back for z, we get two lines:
x-y=0 and x+y-1=0.
Hence, the locus of midpoints is given by the equation:
x2+y2-x-3y=0
Therefore, option 'B' is correct.
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Community Answer
The locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 pa...
The equation of the chord if mid point is given--
T=S1. let midpoint be (h,k)
xh+yk-(x+h)-3(y+k)-10=h2+k2-2h-6k-10
passing through origin (0,0)
-h-3k = h2+k2-2h-6k
h2+k2-h-3k = 0
putting h,k= x,y
x2+y2-x-3y = 0
T=S1. let midpoint be (h,k)
xh+yk-(x+h)-3(y+k)-10=h2+k2-2h-6k-10
passing through origin (0,0)
-h-3k = h2+k2-2h-6k
h2+k2-h-3k = 0
putting h,k= x,y
x2+y2-x-3y = 0
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The locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 passing through the origin isa)x2+y2+x-3y=0b)x2+y2-x-3y=0c)x2+y2-x+3y=0d)x2+y2+x+3y=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 locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 passing through the origin isa)x2+y2+x-3y=0b)x2+y2-x-3y=0c)x2+y2-x+3y=0d)x2+y2+x+3y=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 locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 passing through the origin isa)x2+y2+x-3y=0b)x2+y2-x-3y=0c)x2+y2-x+3y=0d)x2+y2+x+3y=0Correct answer is option 'B'. Can you explain this answer?.
The locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 passing through the origin isa)x2+y2+x-3y=0b)x2+y2-x-3y=0c)x2+y2-x+3y=0d)x2+y2+x+3y=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 locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 passing through the origin isa)x2+y2+x-3y=0b)x2+y2-x-3y=0c)x2+y2-x+3y=0d)x2+y2+x+3y=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 locus of mid point of the chords of the circle x2+y2-2x-6y-10=0 passing through the origin isa)x2+y2+x-3y=0b)x2+y2-x-3y=0c)x2+y2-x+3y=0d)x2+y2+x+3y=0Correct answer is option 'B'. Can you explain this answer?.
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