If OA and OB are two equal chords of the circle x^2+y^2-2x+4y=0 perpendicular to each other and passing through the origin, the slopes of A and OB are the roots of the equation
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p2-2a2
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Given : OA and OB are two equal chords of the circle x^2+y^2-2x+4y=0 perpendicular to each other and passing through the origin,
To Find : the slopes of A and OB are the roots of the equation
Solution:
x² + y² - 2x + 4y = 0
(x - 1)² + ( y + 2)² = 5
Center ( 1 , - 2)
x² + y² - 2x + 4y = 0
0 + 0 - 0 + 0 = 0
Hence ( 0 , 0) origin lies on circle
and both chord passes through Origin
so O = ( 0 , 0)
Both chords are perpendicular to each other
Hence ∠AOB = 90°
=> AB is Diameter
Let say C is center of circle
C is mid point of AB
as OA = OB Hence
OC ⊥ AB
C = ( 1 , - 2)
O = (0 , 0)
Slope of OC = -2
Hence slope of AB = 1/2 as OC ⊥AB
AB passes through C = ( 1 , - 2)
=> Equation of line AB = y + 2 = (1/2)(x - 1)
=> 2y + 4 = x - 1
(x - 1)² + ( y + 2)² = 5
=> (2y + 4)² + ( y + 2)² = 5
=> (y + 2)² )(4 + 1) = 5
=> y + 2 = ±1
=> y = -1 , - 3
x = 3 , - 1
A and B can be ( -1 , - 3) and ( 3 , - 1)
Slope of OA and OB
= ( 3 , - 1/3 )
Sum of slopes = 3 -1/3 = 8/3
Product of slopes = - 1
Quadratic Equation m² - (8/3)m + (-1) = 0
=> 3m² - 8m - 3 = 0
slopes of OA and OB are the roots of the equation 3m² - 8m - 3 = 0
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