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# Fin the equation of the circle circumscribing the triangle formed by the straight lines:x+y=6,2x+y=4 and x+2y=5.

Asked by 1st March 2013, 10:13 AM
Let the triangle be ABC.
Equation of AB: x+y=6         ... (i)
Equation of BC: 2x+y=4       ... (ii)
Equation of CA: x+2y=5       ... (iii)

Solve equations (i) and (ii) to get the vertices of B.
Solve equations (ii) and (iii) to get the vertices of C.
Solve equations (i) and (iii) to get the vertices of A.

Now, we need to find the equation of the circle passing through A, B and C.

Let the equation of the circle be x^2 + y^2 + 2gx + 2fy + c = 0           ... (A)

Substitute the value of x and y for each of the three coordinates in the above equation.
You will get three equations. Solve them to find f, g, c.

You will then get the required equation of the circle by putting these values in equation (A).
Answered by Expert 3rd March 2013, 10:32 PM
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