Question 864960
I see several ways to get the answer. I cannot tell if one of the ways I would use was used in the first answer you got. Maybe the reasoning was through a different way. I can explain one or more ways. However, I am a long-winded explainer, so you may get more explanation that you bargained for.
 
USING GEOMETRY AND ANALYTIC GEOMETRY:
The tangent to a circle is perpendicular to the radius at the point of tangency, so I would draw the sketch like this:
{{{drawing(600,300,-15,5,-5,5,
grid(0),blue(circle(-8,0,4)),
green(line(-15,8.66,5,-2.887)),
green(line(-15,-8.66,5,2.887)),
blue(circle(-8,0,0.1)),
blue(circle(-8,0,0.15)),
green(circle(-6,3.464,0.1)),
blue(triangle(-8,0,0,0,-6,3.464)),
blue(triangle(-8,0,0,0,-6,-3.464)),
locate(-8.15,0.6,C),locate(-6,4,P),
locate(-7,1.9,4),locate(-1.3,0.6,green(theta)),
green(arc(0,0,3,3,-180,-150)),locate(1,0.6,green(theta)),
green(arc(0,0,3,3,-30,0))
locate(0.1,0.6,O),locate(-6,-3.46,Q),
green(line(-6,0,-6,3.464)),
locate(-5.9,2,green(y)),
green(line(-6.15,3.204,-5.89,3.054)),
green(line(-5.74,3.314,-5.89,3.054))
)}}} The tangents, radii, and the x-axis form right triangles OPC and OQC.
The two triangles are congruent, with P(x,y) and Q(x,-y), so I only need to solve OPC.
I know that the length of leg CP is {{{4}}}, and the hypotenuse OC is {{{8}}} .
{{{sin(theta)=4/8=1/2}}}
That tells me that the angle {{{theta}}} measures {{{30^o}}}
and OPC is a 30-60-90 triangle.
From the tangent of that {{{30^o}}} angle you can also get the slopes of the tangents.
{{{tan(30^o)=1/sqrt(3)=(1/sqrt(3))*(sqrt(3)/sqrt(3))=sqrt(3)/(sqrt(3))^2=sqrt(3)/3}}} .
We multiply times {{{(sqrt(3)/sqrt(3))}}} because we do not like to see square roots in denominators,
so m = +- {{{sqrt(3)/3}}} .
Of course, OP is the line with the negative slope, and OQ is the line with the positive slope.
For OP, {{{y=(-sqrt(3)/sqrt(3)))x}}} and
for OQ, {{{y=(sqrt(3)/sqrt(3)))x}}} .
The altitude of that right triangle splits it into two similar right triangles, CPM, and OPM.
As the ratio of short leg to hypotenuse for OPC is {{{4/8=1/2}}}, the ratio is also the same for CPM, and OPM, so
{{{CM=4*(1/2)=2}}} and the x-coordinate of points M and P (and Q) is {{{-8+2=highlight(-6)}}}
For OP, {{{y=(-sqrt(3)/sqrt(3)))x}}}
and for P, {{{y=mx=m*(-6)=(sqrt(3)/sqrt(3))*(-6)=2sqrt(3)}}} .
On the other hand, for Q, {{{y=-2sqrt(3)}}} .
 
WITHOUT MENTIONING GEOMETRY (too much):
OK, we need to know the equation of the circle, and I could call that analytical geometry, but you learn that in algebra 2 too.
The equation of a circle with radius {{{r}}} and center at {{{"( h , k )"}}} is
{{{(x-h)^2+(y-k)^2=r^2}}} .
For the circle in your problem, that would be
{{{(x-(-8))^2+(y-0)^2=4^2}}} ---> {{{(x+8)^2+y^2=16}}} ---> {{{x^2+16x+64+y^2=16}}}
If {{{y=mx}}} is tangent to the circle, there is only one point in common to circle and line,
so {{{system(x^2+16x+64+y^2=16,y=mx)}}} has one and only one solution.
{{{system(x^2+16x+64+y^2=16,y=mx)}}}-->{{{x^2+16x+64+(mx)^2=16}}}-->{{{x^2+16x+64+m^2x^2=16}}}-->{{{(m+1)x^2+16x+64-16=0}}}-->{{{(m+1)x^2+16x+48=0}}}
For what value of {{{m}}} would that quadratic equation have one and only one solution?
A quadratic {{{ax^2+bx+c=0}}} has one and only one real solution when {{{b^2-4ac=0}}} ,
and then {{{x=-b/2a}}}
In the case of {{{(m+1)x^2+16x+48=0}}} ,
{{{b=16}}} , {{{a=m^2+1}}} , and {{{c=48}}} .
It will have one and only one real solution when
{{{16^2-4*(m^2+1)*48=0}}}-->{{{16^2=4*48*(m^2+1)}}}-->{{{16*16=4*3*16*(m^2+1)}}}-->{{{16*cross(16)=4*3*cross(16)*(m^2+1)}}}-->{{{16=12(m^2+1)}}}
So {{{m^2+1=16/12}}}-->{{{m^2+1=4/3}}}-->{{{m^2=4/3-1}}}-->{{{m^2=1/3}}}
m = +-{{{sqrt(1/3)}}}= +- {{{1/sqrt(3)}}}= +- {{{sqrt(3)/3}}} .
In either case, we can use {{{x=-b/2a}}} to find the same {{{x}}} value:
{{{m^2+1=4/3}}} --> {{{highlight(x=(-16)/(m^2+1))}}}
That is the "x coordinate of P in terms of m" as your problem asked for.
Maybe this is the way your teacher/book expected you to solve the problem.
We can calculate further (apparently not required by your problem):
{{{x=-16/((4/3))=-16*(3/4)=-6}}}