The equation of a chord of the circle x^2+y^2+4x-6y=0 is given by x+2y=0. The equation of the circle described on this chord as diameter
Answers
Given : The equation of a chord of the circle x^2+y^2+4x-6y=0 is given by x+2y=0.
To Find : The equation of the circle described on this chord as diameter
Solution:
x²+y²+4x-6y=0
x+2y=0
=> x = - 2y
substitute in circle equation
=> ( - 2y)² + y² + 4(-2y) - 6y = 0
=> 4y² + y² - 8y - 6y = 0
=> 5y² - 14y = 0
=> y(5y - 14) = 0
=> y = 0 , y = 14/5 = 2.8
x = - 2y
=> x = 0 , - 5.6
End points of diameter =
(0, 0) , ( -5.6 , 2,8)
Center = (-2.8 , 1.4)
Diameter = √(5.6)² + (2.8)² = 2.8√5
=> radius = 1.4√5
=> r² = 9.8
The equation of the circle described on this chord as diameter
= (x - (-2.8))² + (y - 1.4)² = 9.8
=> (x + 2.8)² + (y - 1.4)² = 9.8
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