Power of a point proof question

Question

I’m trying to follow this proof: proof

Specifically this part:

The parameters $t_1,t_2$ of possible common points of line $g: \vec x=\vec p+t\vec v$ (through $P$) and circle $c$ can be determined by inserting the parametric equation into the circle's equation: :$(\vec p+t\vec v)^2-r^2=0 \quad \rightarrow \quad t^2+2t\;\vec p\cdot\vec v +\vec p^2-r^2=0 \ .$

Can anyone explain this further? I don’t understand why the parametric equation of a line is being substituted into the equation for a circle.

I already asked a question to clarify the first part: question to clarify circle part

Answer 1

The parametric equation of the line gives all the points by varying the parameter. You get a vector function $\vec x(t)$ of a scalar argument.

The circle is represented by an implicit equation, which is the locus of the vectors such that they cancel a function (namely the squared distance to the center minus the squared radius). You get a scalar function of a vector argument, $c(\vec x)$.

But combining the two, $c(\vec x(t))$, you get a scalar function of a scalar argument and you need to find the roots, which correspond to the intersection(s). From the value(s) of $t$ you get the point(s).

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Four alternatives are possible: implicit or parametric line equation vs. implicit or parametric circle equation. Each form systems of $k$ equations in $k$ unknowns. The given method is simple as it corresponds to $k=1$.

Answer 2

$r$ is the constant scalar geometric mean of all segment vector lengths through P. Start by squaring algebraically and then take the dot product

$$(\vec p + t \vec v )\cdot (\vec p + t \vec v)=r^2 $$ $$=(\vec p \cdot \vec p ) + 2 t\vec p \vec v + (t\vec v \cdot t \vec v) =r^2 $$ $$=(\vec p \cdot \vec p ) + 2 t\vec p \vec v + (t\vec v \cdot t \vec v)=r^2 $$ $$=(\vec p ^2) + 2 t\vec p \vec v + (t^2 1)=r^2 $$

with the last being dot product of two unit vectors