If A and B are two points on the circle x2 + y2 - 4x + 6y - 3=0 which are farthest and nearest respectively from the point (7,2) then (A) A = (2 – 2v2, -3 – 2v2) (B) A = (2 + 2v2,-3+212) (C) B = (2+ 2V2, -3+2V2) (D) B= (2 - 2V2, -3 - 2/2)
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Answer:
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Given : A and B are two points on the circle x2 + y2 - 4x + 6y - 3=0 which are farthest and nearest respectively from the point (7,2)
To Find : Point A ( farthest )
Solution:
x² + y² - 4x + 6y - 3=0
=> (x - 2)² + ( y + 3)² = 4²
Hence center = ( 2 , -3)
Radius = 4
Farthest and nearest point will lie on line joining center and point
Line joining ( 2 , - 3 ) and ( 7 , 2)
y - 2 ={ ( 2 - (-3))/(7 - 2) } (x - 7)
=> y - 2 = x - 7
=> y = x - 5
(x - 2)² + ( y + 3)² = 4²
=> (x - 2)² + ( x - 5 + 3)² = 4²
=> (x - 2)² + (x - 2)² = 16
=> (x - 2)² = 8
=> x - 2 = ± 2√2
=> x = 2 ± 2√2
y = x - 5
=> y = - 3 ± 2√2
Hence two points are ( 2 ± 2√2 , - 3 ± 2√2)
as (2 + 2√2 , - 3 + 2√2 ) lies between ( 2, - 3) and ( 7 , 2)
Nearest point B = (2 + 2√2 , - 3 + 2√2 )
Hence farthest point from ( 7 , 2) is
A = ( 2 - 2√2 , - 3 - 2√2)
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