A circle x^2+y^2-2ax+2y=0cuts the axis of x at a point A and it circumscribes an equilateral triangle with OA as one of the sides, where O is the origin. Find the value of a and the vertices of the equilateral triangle.
Answers
Given : a circle x^2+y^2-2ax+2y=0 cuts off the axis of x at point A and circumscribe an equilateral triangle with OA as one of the side where o is the origin
To Find : the value of A and the vertices of equilateral triangle
Solution:
x² + y² - 2ax + 2y = 0
=> (x - a)² -a² + ( y + 1)² - 1² = 0
=> (x - a)² + ( y + 1)² = a² + 1
Center = ( a , - 1)
radius = √a² + 1
Circle passes through origin
and ( A , 0)
(√a² + 1)² = ( A - a)² + (0 - 1)²
=> a² + 1 = (A - a)² + 1
=> a² = (A - a)²
=> A = 0 , 2a
A = 0 is origin
Hence A = ( 2a , 0 )
OA = 2a
Third vertex = ( x , y)
=> (x - 0)² + ( y - 0)² = (2a)² => x² + y² = 4a²
(x - 2a)² + ( y - 0)² = a² => x² - 4ax + 4a² + y²= 4a²
x² + y² = x² - 4ax + 4a² + y²
=> x = a
x² + y² = 4a²
=> (a)² + y² = 4a²
=> y = ±√3 a
y = ±√3 a
Third vertex = ( a , ±√3 a )
value of A = ( 2a , 0)
vertices of equilateral triangle
(0 , 0) , ( 2a , 0) , ( a , ±√3 a )
As circle circumscribe triangle hence ( a , -√3 a ) lies on circle
(x - a)² + ( y + 1)² = a² + 1
=> (0)² + ( ±√3 a + 1)² = a² + 1
=> 3a² + 1 ± 2√3 a = a² + 1
=> 2a² = 2√3 a
=> a = ±√3
vertices of equilateral triangle
= ( 0 , 0) , ( ±2√3 , 0) , (±√3 , - 3)
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