One of the limit points of the coaxial system | vedclass

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Question

(A) Let $S_1 = x^2 + y^2 - 6x - 6y + 4 = 0$ and $S_2 = x^2 + y^2 - 2x - 4y + 3 = 0$. The radical axis is given by $S_1 - S_2 = 0$,which is $(x^2 + y^2

Vedclass · Accepted Answer

$(-1, 1)$

Vedclass · Answer

(A) Let $S_1 = x^2 + y^2 - 6x - 6y + 4 = 0$ and $S_2 = x^2 + y^2 - 2x - 4y + 3 = 0$. The radical axis is given by $S_1 - S_2 = 0$,which is $(x^2 + y^2 - 6x - 6y + 4) - (x^2 + y^2 - 2x - 4y + 3) = 0$. Simplifying this,we get $-4x - 2y + 1 = 0$,or $4x + 2y - 1 = 0$. The equation of the coaxial system is $S_1 + \lambda(4x + 2y - 1) = 0$,which is $x^2 + y^2 - (6 - 4\lambda)x - (6 - 2\lambda)y + (4 - \lambda) = 0$. The center of these circles is $(3 - 2\lambda, 3 - \lambda)$ and the radius $r$ is given by $r^2 = (3 - 2\lambda)^2 + (3 - \lambda)^2 - (4 - \lambda)$. For limit points,the radius $r = 0$. Thus,$(3 - 2\lambda)^2 + (3 - \lambda)^2 - (4 - \lambda) = 0$. $9 - 12\lambda + 4\lambda^2 + 9 - 6\lambda + \lambda^2 - 4 + \lambda = 0$. $5\lambda^2 - 17\lambda + 14 = 0$. Solving for $\lambda$,we get $(5\lambda - 7)(\lambda - 2) = 0$,so $\lambda = 2$ or $\lambda = 7/5$. For $\lambda = 2$,the center is $(3 - 2(2), 3 - 2) = (-1, 1)$. For $\lambda = 7/5$,the center is $(3 - 14/5, 3 - 7/5) = (1/5, 8/5)$. Comparing with the options,the correct limit point is $(-1, 1)$.

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is:

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