The speed, not velocity of the pencil is constant at $5$ cm/sec. Velocity is a vector and has a direction, speed is the magnitude of velocity. In general, the slope of the curve is given by the derivative. Through a point $(x_0,y_0)$ on the curve, the slope is $m=\frac {dy}{dx}|_{(x_0,y_0)}$ and the tangent line is $y-y_0=m(x-x_0)$. Then a vector along the tangent is $(1,m)$ (of length $\sqrt{1+m^2}$)and we can make it of length $L$ by multiplying by $\frac L{\sqrt{1+m^2}}$ to get $(\frac L{\sqrt{1+m^2}},\frac {mL}{\sqrt{1+m^2}})$
In the case of a circle of radius $10$ centered at the origin, the equation of the top half is $y=\sqrt{100-x^2}$ so $\frac {dy}{dx}=\frac {-x}{\sqrt{100-x^2}}$. If we want the velocity at $(8,6)$ we have $\frac {dy}{dx}=\frac {-8}{\sqrt{100-64}}=-\frac 43$. The vector along the tangent is then $(-1,\frac 43)$ (note we have to resolve the direction, I assume you are drawing counterclockwise), the length of that vector is $\sqrt{1+\frac {16}9}=\frac 53$, so the velocity is $\dfrac 5{\frac 53}(-1,\frac 43)=(-3,4)$