Why does the circle intersect the line?
Extrapolate the line until it it intersects both coordinate axes. The points of intersection are then $(15/2,0)$ and $(0,-10)$.
The line and the exes now make a right triangle whose legs are $15/2$ and $10$. Now, apply the Pythagorean Theorem, find that the hypoteneuse is $25/2$ (proportional to a 3-4-5 right triangle).
In a right triangle, the area is half the product of the legs and also half the product of the hypoteneuse and the altitude to that hypoteneuse. So the products must be equal and we must have $(10)(15/2)=(25/2)x$ where $x$ is the altitude to the hypoteneuse. Then $x=6$ matching the radius of the given circle; the line is tangent to the circle.
The circle intersects the line if the radius of the circle is at least as large as the distance from the circle's center to the line.
We measure the distance from a point to a line in a direction perpendicular to the line. Seeing that the slope of the line from $C$ to $D$ is $\frac43,$ we want a line from the center of the circle whose slope is $-\frac34,$ that is, the line $y = -\frac34 x.$
Now we lay out the radius of the circle along that line. The radius is $6.$ Since the hypotenuse of a triangle with legs $3$ and $4$ is $5$, we can scale up one such triangle by the factor $\frac65$ to find the desired point on the line $y = -\frac34 x.$ That triangle has legs $\frac{18}{5}$ and $\frac{24}{5},$ and hypotenuse $6,$ which means the point we are seeking is $\left(\frac{24}{5}, -\frac{18}{5} \right).$
If that point is on the line or on the opposite side of the line from the center of the circle, then the circle intersects the line.
As it turns out, the point $\left(\frac{24}{5}, -\frac{18}{5} \right)$ is exactly on the line through $C$ and $D,$ which you can show using geometry without calculus.
A possibly more obvious approach is to find the equation of the line and then compute its distance from the center of the circle.
There are standard formulas for that sort of thing, but if you do not recall the distance formula it is easy enough to derive by building a right triangle with its right angle at the center of the circle, its hypotenuse on the line, and its legs parallel to the axes. (In this case, in fact, its legs are on the axes.)
Find the area of the triangle two ways: using the lengths of the legs, and using the length of the hypotenuse and the altitude from the right-angle vertex. The areas must be the same, of course, since it's the same triangle. Solve for the altitude, which is the distance to the line.
And yet another alternative:
Find the area of the triangle $\triangle ACD.$ Using this area and the fact that $CD = 5,$ find the altitude of $\triangle ACD$ from vertex $A.$ If the altitude is no greater than $6,$ the circle intersects the line.
The line through C and D has equation $$ 4x-3y-30=0 $$ The distance from the line to the point $(0,0)$ is given by the formula $$ d=\frac{|4(0)-3(0)-30|}{\sqrt{4^2 + (-3)^2}}$$ If $d \le 6$ we have intersection.