Question

The circles, which cut the family of circles passing through the fixed points A = (2, 1) and B = (4,3) orthogonally, pass through two fixed points $(x_1, y_1)$ and $(x_2, y_2)$, which may be real or non real. Find the value of $(x_1^3 + x_2^3 + y_1^3 + y_2^3)$.

• A. 260
• B. 116
• C. 144
• D. 200

Answer 1

Answer: (A)

260

Answer 2

The family of circles passing through A(2,1) and B(4,3) can be represented as $S + \lambda L = 0$, where $S=0$ is the circle with AB as diameter and $L=0$ is the line AB. The line AB has equation: $y-1 = \frac{3-1}{4-2}(x-2) \implies y-1 = \frac{2}{2}(x-2) \implies y-1 = x-2 \implies x-y-1=0$. The circle with AB as diameter has center $(\frac{2+4}{2}, \frac{1+3}{2}) = (3,2)$ and radius squared $r^2 = (\frac{4-2}{2})^2 + (\frac{3-1}{2})^2 = 1^2+1^2=2$. Its equation is $(x-3)^2 + (y-2)^2 = 2 \implies x^2-6x+9+y^2-4y+4=2 \implies x^2+y^2-6x-4y+11=0$. So, the first family of circles is $x^2+y^2-6x-4y+11 + \lambda(x-y-1) = 0$, which is $x^2+y^2+(-6+\lambda)x+(-4-\lambda)y+(11-\lambda)=0$. Let the orthogonal family of circles be $x^2+y^2+2g_2x+2f_2y+c_2=0$. The orthogonality condition $2g_1g_2+2f_1f_2=c_1+c_2$ for all $\lambda$ implies: $2(\frac{-6+\lambda}{2})g_2 + 2(\frac{-4-\lambda}{2})f_2 = (11-\lambda) + c_2$ $(-6+\lambda)g_2 + (-4-\lambda)f_2 = 11-\lambda+c_2$ $\lambda(g_2-f_2+1) = 11+6g_2+4f_2-c_2$. For this to hold for all $\lambda$:

1. $g_2-f_2+1=0 \implies f_2 = g_2+1$.
2. $11+6g_2+4f_2-c_2=0 \implies 11+6g_2+4(g_2+1)-c_2=0 \implies 11+6g_2+4g_2+4-c_2=0 \implies 15+10g_2-c_2=0 \implies c_2=10g_2+15$. The orthogonal family of circles is $x^2+y^2+2g_2x+2(g_2+1)y+(10g_2+15)=0$. This can be written as $(x^2+y^2+2y+15) + g_2(2x+2y+10) = 0$. The fixed points are the intersections of the radical axis $2x+2y+10=0$ (or $x+y+5=0$) with any circle in the family. Let's use $x^2+y^2+2y+15=0$ (when $g_2=0$). From $x+y+5=0$, we have $y=-x-5$. Substitute this into the circle equation: $x^2+(-x-5)^2+2(-x-5)+15=0$ $x^2+(x^2+10x+25)-2x-10+15=0$ $2x^2+8x+30=0 \implies x^2+4x+15=0$. This quadratic gives the x-coordinates $x_1, x_2$. By Vieta's formulas: $x_1+x_2=-4$ and $x_1x_2=15$. Since $y_1=-x_1-5$ and $y_2=-x_2-5$: $y_1+y_2 = -(x_1+x_2)-10 = -(-4)-10 = 4-10 = -6$. $y_1y_2 = (-x_1-5)(-x_2-5) = (x_1+5)(x_2+5) = x_1x_2+5(x_1+x_2)+25 = 15+5(-4)+25 = 15-20+25=20$. We need to calculate $x_1^3+x_2^3+y_1^3+y_2^3$. Using the identity $a^3+b^3=(a+b)^3-3ab(a+b)$: $x_1^3+x_2^3 = (x_1+x_2)^3-3x_1x_2(x_1+x_2) = (-4)^3-3(15)(-4) = -64+180 = 116$. $y_1^3+y_2^3 = (y_1+y_2)^3-3y_1y_2(y_1+y_2) = (-6)^3-3(20)(-6) = -216+360 = 144$. The total sum is $116+144=260$. The roots of $x^2+4x+15=0$ are $x = \frac{-4 \pm \sqrt{16-60}}{2} = -2 \pm i\sqrt{11}$, which are non-real as stated in the problem.