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The equation of the circle having the lines y2 – 2y + 4x – 2xy = 0 as its normals & passing through the point (2, 1) is
- a)x2 + y2 – 2x – 4y + 3 = 0
- b)x2 + y2 – 2x + 4y – 5 = 0
- c)x2 + y2 + 2x + 4y – 13 = 0
- d)None
Correct answer is option 'A'. Can you explain this answer?
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The equation of the circle having the lines y2–2y + 4x –2x...
The normal line to circle is →y² - 2 y + 4 x -2 xy=0
→ y(y-2) - 2x(y-2)=0
→ (y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)² + (2-1)²]½
=(1+1)½ =(2)½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→ (x-1)²+(y-2)²= 2
x²+y² -2x-4y+3 = 0
→ y(y-2) - 2x(y-2)=0
→ (y-2)(y-2x)=0
the two lines are , y=2 and 2 x -y =0
The point of intersection of normals are centre of circle.
→ Put , y=2 in 2 x -y=0, we get
→2 x -2=0
→2 x=2
→ x=1
So, the point of intersection of normals is (1,2) which is the center of circle.
Also, the circle passes through (2,1).
Radius of circle is given by distance formula = [(1-2)² + (2-1)²]½
=(1+1)½ =(2)½
The equation of circle having center (1,2) and radius √2 is
= (x-1)²+(y-2)²=[√2]²
→ (x-1)²+(y-2)²= 2
x²+y² -2x-4y+3 = 0
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The equation of the circle having the lines y2–2y + 4x –2xy = 0 as its normals & passing through the point (2, 1) isa)x2+ y2–2x –4y + 3 = 0b)x2+ y2–2x + 4y –5 = 0c)x2+ y2+ 2x + 4y –13 = 0d)NoneCorrect answer is option 'A'. Can you explain this answer?
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The equation of the circle having the lines y2–2y + 4x –2xy = 0 as its normals & passing through the point (2, 1) isa)x2+ y2–2x –4y + 3 = 0b)x2+ y2–2x + 4y –5 = 0c)x2+ y2+ 2x + 4y –13 = 0d)NoneCorrect answer is option 'A'. 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 having the lines y2–2y + 4x –2xy = 0 as its normals & passing through the point (2, 1) isa)x2+ y2–2x –4y + 3 = 0b)x2+ y2–2x + 4y –5 = 0c)x2+ y2+ 2x + 4y –13 = 0d)NoneCorrect answer is option 'A'. 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 having the lines y2–2y + 4x –2xy = 0 as its normals & passing through the point (2, 1) isa)x2+ y2–2x –4y + 3 = 0b)x2+ y2–2x + 4y –5 = 0c)x2+ y2+ 2x + 4y –13 = 0d)NoneCorrect answer is option 'A'. Can you explain this answer?.
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