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Find the equation of the circle which passes through the points (1, 2), (2, 2) and (4, -1)?

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square of (x-3/2) + square of (y-3/2) = 1/2.
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Take the general equation for a circle: x^2 + y^2 + 2gx + 2fy + c = 0, where the center is at (-g, -f) and the radius is sq root of (g^2 +f^2 -c). Now use the given points to get three equations to solve for three unknowns, g, f, and c. The equations with point (1, 2) becomes: 1^2 + 2^2 + 2xgx1 + 2xfx2...
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Take the general equation for a circle: x^2 + y^2 + 2gx + 2fy + c=0, where the center is at (-g, -f) and the radius is sq root of (g^2 +f^2 -c). Now use the given points to get three equations to solve for three unknowns, g, f, and c. The equations with point (1, 2) becomes: 1^2 + 2^2 + 2xgx1 + 2xfx2 + c=0 => 2g + 4f + c=- 5 .............. (1) Similarly, the equation with the point (2, 2) is: => 4g + 4f + c=- 8 .............. (2) And, the equation with the point (4, -1) is: => 8g - 2f + c=- 17 .............. (3) From (1) - (2) we get: -2g = 3 --> g=-3/2. Using this value of g in (1) we get: 4f + c=- 2 ............ (4) Using this value of g in (3) we get: -2f + c=- 5 ............ (5) From (4) - (5) we can get: 6f=3 --> f=1/2. Using this value of f in (5) we get: c=- 4. So, the equation of the circle is: x^2 + y^2 + 2(-3/2)x + 2(1/2)y + (- 4) = 0. After simplification: --> x^2 + y^2 -3x +y - 4 = 0 (Answer) read less
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