analytic geometrycircleslinear algebravectors
If the equation for a circle is $|c-x|^2 = r^2$ and the equation for the line is $n \cdot x=d $, and assuming that the circle and line intersect in two points, how can I find these points?
Also as kind of a side note, is there also a general formula for obtaining the equation of a circle of the intersection of a plane and sphere in $\mathbb R^3$, using similar equations?
Let's say you have the line $y = mx + c$ and the circle $(x-p)^2 + (y-q)^2 = r^2$.
First, substitute $y = mx + c$ into $(x-p)^2 + (y-q)^2 = r^2$ to give
$$(x-p)^2 + (mx+c-q)^2 = r^2 \, . $$
Next, expand out both brackets, bring the $r^2$ over to the left, and collect like terms:
$$(m^2+1)x^2 + 2(mc-mq-p)x + (q^2-r^2+p^2-2cq+c^2) = 0 \, .$$
This is a quadratic in $x$ and can be solved using the quadratic formula. Let us relabel the coefficients to give $Ax^2 + Bx + C = 0$, then we have
$$x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} \, . $$
If $B^2-4AC < 0$ then the line misses the circle. If $B^2-4AC=0$ then the line is tangent to the circle. If $B^2-4AC > 0$ then the line meets the circle in two distinct points.
Since $y=mx+c$, we can see that if $x$ is as above then
$$y = m\left(\frac{-B \pm \sqrt{B^2-4AC}}{2A}\right) + c \, . $$
EDIT: The lines $y=mx+c$ do not cover the vertical lines $x=k$. In that case, substitute $x=k$ into $(x-p)^2+(y-q)^2=r^2$ to give
$$y^2 - 2qy + (p^2+q^2-r^2 - 2kp+k^2) = 0$$
This gives a quadratic in $y$, namely $y^2+By+C=0$, where $B=-2q$ and $C=p^2+q^2-r^2 - 2kp+k^2$. Solve using the Quadratic Formula. The solutions are $(k,y_1)$ and $(k,y_2)$, where the $y_i$ are solutions to $y^2+By+C=0$.
$\newcommand{\Vec}[1]{\mathbf{#1}}$Generalities: Let $S$ be the sphere in $\mathbf{R}^{3}$ with center $\Vec{c}_{0} = (x_{0}, y_{0}, z_{0})$ and radius $R > 0$, and let $P$ be the plane with equation $Ax + By + Cz = D$, so that $\Vec{n} = (A, B, C)$ is a normal vector of $P$.
If $\Vec{p}_{0}$ is an arbitrary point on $P$, the signed distance from the center of the sphere $\Vec{c}_{0}$ to the plane $P$ is $$ \rho = \frac{(\Vec{c}_{0} - \Vec{p}_{0}) \cdot \Vec{n}}{\|\Vec{n}\|} = \frac{Ax_{0} + By_{0} + Cz_{0} - D}{\sqrt{A^{2} + B^{2} + C^{2}}}. $$
The intersection $S \cap P$ is a circle if and only if $-R < \rho < R$, and in that case, the circle has radius $r = \sqrt{R^{2} - \rho^{2}}$ and center $$ \Vec{c} = \Vec{c}_{0} + \rho\, \frac{\Vec{n}}{\|\Vec{n}\|} = (x_{0}, y_{0}, z_{0}) + \rho\, \frac{(A, B, C)}{\sqrt{A^{2} + B^{2} + C^{2}}}. $$
Now consider the specific example $$ S = \{(x, y, z) : x^{2} + y^{2} + z^{2} = 4\},\qquad P = \{(x, y, z) : x - z\sqrt{3} = 0\}. $$ The center of $S$ is the origin, which lies on $P$, so the intersection is a circle of radius $2$, the same radius as $S$.
When you substitute $x = z\sqrt{3}$ or $z = x/\sqrt{3}$ into the equation of $S$, you obtain the equation of a cylinder with elliptical cross section (as noted in the OP). However, you must also retain the equation of $P$ in your system. That is, each of the following pairs of equations defines the same circle in space: \begin{align*} x - z\sqrt{3} &= 0, & x - z\sqrt{3} &= 0, & x - z\sqrt{3} &= 0, \\ x^{2} + y^{2} + z^{2} &= 4; & \tfrac{4}{3} x^{2} + y^{2} &= 4; & y^{2} + 4z^{2} &= 4. \end{align*} These may not "look like" circles at first glance, but that's because the circle is not parallel to a coordinate plane; instead, it casts elliptical "shadows" in the $(x, y)$- and $(y, z)$-planes.
Note that a circle in space doesn't have a single equation in the sense you're asking.
Best Answer
The answer given by Seyhmus is fine if the line is a line through the origin (or $d = 0$ in your formula). Perhaps a better solution, in the general case, is to find the interserction of a circle $|c - x|^2 = r^2$ with a parametric line, $$ x(t) = a + tb $$ where $a$ is a point and $b$ is a vector.For a point on this line to satisfy the equation, you need to have $$ (tb + (a-c)) \cdot (tb + (a-c)) = r^2 $$ which is a quadratic in $t$: $$ |b|^2 t^2 + 2(a-c)\cdot b t + (|a-c|^2 - r^2) = 0 $$ whose solutions are $$ t = \frac{-2(a-c)\cdot b \pm \sqrt{[2(a-c)\cdot b]^2 - 4|b|^2(|a-c|^2 - r^2)}}{2|b|^2} $$ Once you find these two values of $t$ and plug them back intot he parametric equation, you get your two points.
Coincidentally, this method works fine in all dimensions (i.e., $c$ and $a$ can be points of $R^3$, and $b$ a vector in $R^3$), and the same formula works.
In two dimensions, it's pretty easy to convert from the implicit form of the line to the parametric form: find any two points $P$ and $Q$ on the line, and let $a = P$ and $b = (Q - P)$. In three dimensions, it's rather difficult to specify a line implicitly, but quite simple to express it parametrically, so the general case tends to be asked in exactly the form in which I've answered it.