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. Since the radical axis has a slope of $-\frac{1}{2}$,the line of centres has a slope of $2$. Thus,the equation of the line of centres is of the form $2x - y + k = 0$. The base circle $S_1 = x^2 + y^2 - 3x + \frac{3}{2}y - \frac{3}{4} = 0$ has its centre at $(\frac{3}{2}, -\frac{3}{4})$. Since this centre must lie on the line of centres,we substitute the coordinates into $2x - y + k = 0$: $2(\frac{3}{2}) - (-\frac{3}{4}) + k = 0$ $3 + \frac{3}{4} + k = 0$ $k = -\frac{15}{4}$. Substituting $k$ back into the equation,we get $2x - y - \frac{15}{4} = 0$,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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