If x^2+y^2-4x-2y+5=0 and x^2+y^2-6x-4y-3=0 ar | vedclass

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(A) The equation of the coaxial system of circles is given by $S_1 + \lambda(S_1 - S_2) = 0$,where $S_1 = x^2+y^2-4x-2y+5$ and $S_2 = x^2+y^2-6x-4y-3$

Vedclass · Accepted Answer

$(-5, -6)$

Vedclass · Answer

(A) The equation of the coaxial system of circles is given by $S_1 + \lambda(S_1 - S_2) = 0$,where $S_1 = x^2+y^2-4x-2y+5$ and $S_2 = x^2+y^2-6x-4y-3$.First,find the radical axis: $S_1 - S_2 = (x^2+y^2-4x-2y+5) - (x^2+y^2-6x-4y-3) = 2x + 2y + 8 = 0$,which simplifies to $x + y + 4 = 0$.The family of circles is $x^2+y^2-4x-2y+5 + \lambda(x+y+4) = 0$.$x^2+y^2 + x(\lambda-4) + y(\lambda-2) + (4\lambda+5) = 0$.The centre of these circles is $(-\frac{\lambda-4}{2}, -\frac{\lambda-2}{2})$.$A$ point circle has radius $r = 0$,so $g^2 + f^2 - c = 0$.$(\frac{\lambda-4}{2})^2 + (\frac{\lambda-2}{2})^2 - (4\lambda+5) = 0$.$\frac{\lambda^2-8\lambda+16 + \lambda^2-4\lambda+4}{4} - 4\lambda - 5 = 0$.$2\lambda^2 - 12\lambda + 20 - 16\lambda - 20 = 0$.$2\lambda^2 - 28\lambda = 0 \implies 2\lambda(\lambda - 14) = 0$.For $\lambda = 14$,the centre is $(-\frac{14-4}{2}, -\frac{14-2}{2}) = (-5, -6)$.

If $x^2+y^2-4x-2y+5=0$ and $x^2+y^2-6x-4y-3=0$ are members of a coaxial system of circles,then the centre of a point circle in the system is

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