You can represent the 3d circle in parametric form:
1) form a local coordinate system X'Y'Z' on the plane with origin at the circle's center and Z' axis in the same direction as plane's normal.
2) This 3d circle can be represented as
$(x'(t), y'(t), z'(t)) = (rcos(t), rsin(t), 0)$ where r=radius
3) Perform coordinate transformation between coordinate systems X'Y'Z' and XYZ.
I will leave the details about how to find the local coordinate system and how to perform coordinate transformation to yourself.
Center of circle: at $(0,0,3)$ , radius = $3$.
Plane $z = x +3$ passes through center.
Normal of plane: $(1,0,-1)$ .
Let vector $(a,b,c)$ be perpendicular to this normal:
$(a,b,c) \cdot (1,0,-1)$ = $0$ ; $a - c = 0$.
Yields 2 independent, orthogonal vectors perpendicular to the normal $(1,0,-1)$ of the plane:
1) $(1/√2)(1,0,1)$ and
2) $(0,1,0)$.
Let $\vec{s}$ = $\alpha (1/√2)(1,0,1) +\beta (0,1,0)$.
In analogy to a circle traced in the $x, y$ - plane:
$\vec{s} \cdot (1/√2)(1,0,1)$ = $3 cos(\theta)$ = $\alpha$.
$\vec{s} \cdot (0,1,0)$ = $3 sin(\theta)$ = $\beta$
Finally the parameter representation of the great circle:
$\vec{r}$ = $(0,0,3) + (1/√2)3cos(\theta)(1,0,1) + 3sin(\theta)(0,1,0)$
Best Answer
We determine the distance of the points $C(5/-3/16)$ and $(5/3/16)$ , which is $6$ and denote it with $d$ ( d is the distance of $C$ from the plane). Then, the radius of the circle is $r'=\sqrt{r^2-d^2}=\sqrt{14^2-6^2}=\sqrt{160}=12.649$