find the equation of circle whose diameter is common chord of circles x^2 + y^2 + 2x + 3y + 1=0 and
x^2 + y^2 + 4x + 3y + 2 =0
ans : - 2x ^2 + 2y^2 -2x -6y -1 =0
Center of the circl is (-1,-3/2) and (-2,-3/2)
Means The line passing through centre of both the circle can be written as y=-3/2
and this line will divide the chord in middle point whose y co-ordinate will be -3/2.
And since the eqution of chord is 2x+1=0
so x-cprdinate will be -1/2
So finally without hitting your mind you get the centre of the circle....as (-1/2 , -3/2)
Now to find the radius...simply apply pythagorus theorem....
Perpendicular distance from (-1.-3/2) to 2x+1=0
and radius of first circle...to find the half distance of chord whic is the radius of circle which we are trying to find.........
In anurag solution.......the two end points of chord which he has calculated........is right...and when you take the middle point of the two end point you will get the same point (-1/2.-3/2)
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