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A triangle has two of its sides lying along the x-axis and the line x + y + 1 = 0. The orthocenter of the triangle is at the point (1, 1). Find the equation of the circle that passes through all three vertices of the triangle.
- x² + y² - 3x + y = 0
- x² + y² + x + 3y = 0
- x² + y² + 2y - 1 = 0
- x² + y² + x + y = 0
Correct answer: x² + y² + x + 3y = 0
Solution
The vertices come out to A=(-1,0), B=(0,0), C=(1,-2). Substituting into x²+y²+Dx+Ey+F=0 gives D=1, E=3, F=0, yielding x²+y²+x+3y=0.
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