Find the equation of the circle passing through the vertices of a triangle whose sides are represented by the equations x + y - 2 = 0, 3x - 4y - 6 = 0 and x - y = 0.
Answers
x² + y² + 4x + 6y - 12 = 0 or (x + 2)² + (y + 3)² = 5² is equation of the circle passing through the vertices of a triangle whose sides are represented by the equations x + y - 2 = 0, 3x - 4y - 6 = 0 and x - y = 0
Step-by-step explanation:
x + y - 2 = 0, Eq1
3x - 4y - 6 = 0 Eq 2
x - y = 0. Eq 3
Eq1 + Eq 3
=> 2x - 2 = 0 => x = 1 => y = 1
(1 , 1)
4*eq1 + Eq2
=> 7x - 14 = 0 => x = 2 => y = 0
(2 , 0)
Eq2 - 4 * eq3
=> -x - 6 = 0 => x = - 6 => y = -6
(-6 , - 6)
Equation of circle passing through
(1 , 1) , ( 2, 0) & (-6 , - 6)
x² + y² + 2gx + 2fy + c = 0
=> 1 + 1 + 2g + 2f + c = 0
=> 2g + 2f + c = - 2 EqA
4 + 0 + 4g + 0 + c = 0
=> 4g + c = - 4 EqB
36 + 36 - 12g - 12f + c = 0
=> 12g + 12f - c = 72 EqC
6*EqA - Eq C
=> 7c = -84
=> c = -12
4g + c = - 4
=> 4g - 12 = -4
=> 4g = 8
=> g = 2
2g + 2f + c = - 2
4 + 2f - 12 = -2
=> 2f = 6
=> f = 3
x² + y² + 4x + 6y - 12 = 0
(x + 2)² + (y + 3)² = 5²
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