The equation of the line joining the centres | vedclass

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(A) The given coaxial system of circles is $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$.Dividing by $4$,we get $x^2 + y^2 - 3x + \frac{3}{2}

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$8x - 4y - 15 = 0$

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(A) The given coaxial system of circles is $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$.Dividing by $4$,we get $x^2 + y^2 - 3x + \frac{3}{2}y - \frac{3}{4} + \frac{\lambda}{4}(x + 2y - 6) = 0$.The radical axis is $x + 2y - 6 = 0$. The line of centres is perpendicular to the radical axis,so it is of the form $2x - y + k = 0$.The centre of the base circle $x^2 + y^2 - 3x + \frac{3}{2}y - \frac{3}{4} = 0$ is $(\frac{3}{2}, -\frac{3}{4})$.Since the line of centres passes through this point,we substitute it into $2x - y + k = 0$:$2(\frac{3}{2}) - (-\frac{3}{4}) + k = 0 \implies 3 + \frac{3}{4} + k = 0 \implies k = -\frac{15}{4}$.Thus,the equation of the line of centres is $2x - y - \frac{15}{4} = 0$,which simplifies to $8x - 4y - 15 = 0$.

The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

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