Circle 1: centre $C_1=(2,3)$, $r_1=\sqrt{4+9+12}=5$.

Circle 2: centre $C_2=(-3,-9)$, $r_2=\sqrt{9+81-26}=8$.

Distance $|C_1C_2|=\sqrt{(2+3)^2+(3+9)^2}=\sqrt{25+144}=\sqrt{169}=13$.

Since $|C_1C_2|=13=r_1+r_2=5+8$, the circles are externally tangent. Externally tangent circles have exactly 3 common tangents.

Rewrite the expression to have a denominator that is a power of 10:

$\dfrac{1}{2^{11} \cdot 5^{17}} = \dfrac{1}{2^{11} \cdot 5^{17}} \times \dfrac{2^6}{2^6} = \dfrac{2^6}{2^{17} \cdot 5^{17}} = \dfrac{64}{10^{17}}$

So the decimal is $\dfrac{64}{10^{17}} = 0.\underbrace{00\ldots0}_{15}64$

The nonzero digits are $6$ and $4$ — that is, two nonzero digits.

Split into right and left limits.

Right limit ($x\to0^+$): $[x]=0$, so $\sin(x[x])=0$, $|x|=x$.

$\dfrac{\tan(\pi\sin^2x)+x^2}{x^2}$. Using $\tan(\pi\sin^2x)\approx\pi\sin^2x\approx\pi x^2$:

$=\pi+1$

Left limit ($x\to0^-$): $[x]=-1$, $\sin(x[x])=\sin(-x)=-\sin x$, $|x|=-x$.

$=\dfrac{\tan(\pi\sin^2x)+(-x-(-\sin x))^2}{x^2}=\dfrac{\pi x^2+(\sin x-x)^2}{x^2}\to\pi+0=\pi$

Left $\neq$ Right, so the limit does not exist.

Convert both distances to the same unit.

Runner A ran $\dfrac{4}{5}$ km $= 0.8$ km $= 800$ meters.

Runner B ran $800$ meters.

Both runners covered exactly the same distance: 800 meters.

Therefore the two quantities are equal.