The point (2, 3) is a limiting point of a coa | vedclass

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(A) The equation of a coaxial system of circles is given by $S + \lambda L = 0$,where $S = x^2 + y^2 - 9 = 0$ and $L$ is the radical axis.Since $(2, 3

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$\left( \frac{18}{13}, \frac{27}{13} \right)$

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(A) The equation of a coaxial system of circles is given by $S + \lambda L = 0$,where $S = x^2 + y^2 - 9 = 0$ and $L$ is the radical axis.Since $(2, 3)$ is a limiting point,the circle with radius $0$ passing through $(2, 3)$ is $(x-2)^2 + (y-3)^2 = 0$,which is $x^2 + y^2 - 4x - 6y + 13 = 0$.The radical axis is $S_1 - S_2 = 0$,which gives $(x^2 + y^2 - 9) - (x^2 + y^2 - 4x - 6y + 13) = 0$,simplifying to $4x + 6y - 22 = 0$ or $2x + 3y - 11 = 0$.The family of coaxial circles is $(x^2 + y^2 - 9) + k(2x + 3y - 11) = 0$.The center is $(-k, -\frac{3k}{2})$ and the radius squared is $r^2 = k^2 + \frac{9k^2}{4} + 9 + 11k = \frac{13k^2 + 44k + 36}{4}$.For limiting points,$r^2 = 0$,so $13k^2 + 44k + 36 = 0$.Solving for $k$,we get $(13k + 18)(k + 2) = 0$,so $k = -2$ or $k = -\frac{18}{13}$.For $k = -2$,the center is $(2, 3)$.For $k = -\frac{18}{13}$,the center is $(\frac{18}{13}, \frac{3}{2} 	imes \frac{18}{13}) = (\frac{18}{13}, \frac{27}{13})$.

The point $(2, 3)$ is a limiting point of a coaxial system of circles of which $x^2 + y^2 = 9$ is a member. The coordinates of the other limiting point are given by

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