Find the intersection points of two circles Find the intersection points of the circles $$k_1:(x-4)^2+(y-1)^2=9\\k_2:(x-8)^2+(y+4)^2=100$$

The intersections point (if they exist) will satisfy the equations of both the circles, so we can find their coordinates by solving the system $$\begin{cases}(x-4)^2+(y-1)^2=9\\(x-8)^2+(y+4)^2=100\end{cases}\iff\begin{cases}x^2-8x+y^2-2y=-8\\x^2-16x+y^2+8y=20\end{cases}$$
Substracting these equations, gives $4x-5y=-14\Rightarrow x=\dfrac{5y-14}{4}$. Substituting into the first, I got (if I didn't mess up the calculations) $$41y^2-332y+766=0$$ which has no real solutions (negative discriminant). Therefore the system has no solutions as well, so the circles don't intersect. Is there something else we can use in order to conclude that they do not intersect? Maybe something which requires less calculations? The center of the first is $O_1(4;1)$, the center of the second cirlce is $O_2(8;-4)$ and their radii are $r_1=3$ and $r_2=10$, respectively, if this somehow helps. The distance $O_1O_2=\sqrt{16+25}=\sqrt{41}$ and $r_1+r_2=13=\sqrt{169}$, but I am not sure how to interpret these findings.

Geometric point of view: The radical axis of two circles passes through the intersection points of the circles when they intersect. If we show that the distance of the centers of the circles to that line is larger than their radii, we're done.
You found the radical axis. It is $L:4x-5y+14=0$. Now, $h_1=d(L,O_1)=\frac{|4\times4-5\times1+14|}{\sqrt{4^2+5^2}}=\frac{25}{\sqrt{41}}>\frac{25}{7}>3=R_1$ and similarly $h_2=d(L,O_2)==\frac{|4\times8-5\times(-4)+14|}{\sqrt{4^2+5^2}}=\frac{66}{\sqrt{41}}>10=R_2$, since $66^2=4356>4100.$
