Radical Axis

@Marvin Kalngan , thanks for this problem, I learned what radical axis is.

Referring to Wikipedia , I noted that the radical axis is always perpendicular to the line connecting the centers of the two circles involved. In this case they are $C_1 : (x-1)^2 + y^2 = 3$ and $C_2: (x+2)^2 + (y-4)^2 = 8$ with centers at $O_1(1,0)$ and $O_2(-2,4)$ respectively. Then the gradient of $O_1O_2$ is $\frac {4-0}{-2-1} = -\frac 43$ . Therefore, the gradient $m$ of the radical axis is $-\frac 43m = -1$ , $\implies m = \frac 34$ , and its equation is $y = mx + c = \frac 34 x + c \implies 3x - 4y = k$ , where $k$ is a constant. Since there is only one answer option of this form, the answer must be $\boxed{3x-4y=-7}$ .

So, Marvin, we shouldn't set such problem as an objective question. We can ask for, for example, "The equation of the radical axis is of the form $ax+by=c$ . Find $a+b+c$ ."

To find $k$ of the equation of the radical axis, we use another property of radical axis. Each point $P$ on the radical axis has the same power with respect to both circles. And there is a circle centered at $P$ with radius $R$ such that $R^2 = d_1^2 - r_1^2 = d_2^2-r_2^2$ , where $r_1$ and $r_2$ are the radii of the two circles and $d_1$ and $d_2$ are the distances between center $P$ and the respective centers of the two circles.

Let the coordinates of $P$ be $(x,y)$ . Then we have

$\begin{aligned} d_1^2 - r_1^2 & = d_2 - r_2^2 \\ (x-1)^2 + (y-0)^2 - 3 & = (x+2)^2 + (y-4)^2 - 8 \\ x^2 - 2x + 1 + y^2 - 3 & = x^2 + 4x + 4 + y^2 - 8y + 16 - 8 \\ 6x - 8y & = -14 \\ 3x - 4y & = -7 \Longleftarrow \boxed{\text{The answer}} \end{aligned}$

Solution is clearly given in the link attached to the problem. The equation of the radical axis is $2(1+2)x+2(0-4)y=8-3+(1)^2+(0)^2-(-2)^2-(4)^2\implies \boxed {3x-4y=-7}$ .