Since $a^{2}\sin θ+a\cos θ- \frac {π}{4}=0$, $b^{2}\sin θ+b\cos θ- \frac {π}{4}=0$, we have $\begin{cases} \cos θ= \frac {π(a+b)}{4ab} \ \sin θ=- \frac {π}{4ab}\end{cases}$

Given that $\sin ^{2}θ+\cos ^{2}θ=1$, we get $\frac {ab}{ \sqrt {1+(a+b)^{2}}}= \frac {π}{4}$

The equation of the line passing through the two points $((a,a^{2})$, $(b,b^{2}))$ is $((b+a)x-y-ab=0)$

The expression $\frac {ab}{ \sqrt {1+(a+b)^{2}}}= \frac {π}{4}$ indicates that the distance between $(0,0)$ and $((b+a)x-y-ab=0)$ is $\frac {π}{4}$

Therefore, the line intersects with the circle $x^{2}+y^{2}=1$.

Hence, the answer is $\boxed{\text{C}}$.

Using the given equations, we find $\sin θ$ and $\cos θ$; using the square relationship of trigonometric functions, we get the equation that $a$ and $b$ satisfy; using the two-point form, we find the equation of the line, and using the formula for the distance between a point and a line and the condition for a line to be tangent to a circle, we find the equation of the circle.

This problem tests the relationship between a line and a circle, mainly examining the square relationship of trigonometric functions, the two-point form to find the equation of a line, the formula for the distance between a point and a line, and the condition for a line to be tangent to a circle.