SOLUTION: I previously asked this question and got this solution (after question). But I would appreciate further clarification. The line with the equation y=mx is tangent to the circle wit

Question 864960: I previously asked this question and got this solution (after question). But I would appreciate further clarification.
The line with the equation y=mx is tangent to the circle with centre (-8,0) and radius 4 at the point P(x,y).�
a)find x coordinate of P in terms of m�
b)Show that m = +- sqrt (3)/ 3 and hence find the coordinates of P
(x+8)^2+y^2=16
(x+8)^2+y^2=16, y=mx
m = 1/sqrt(3),�� x = -6,�� y = m x
m =-1/sqrt(3),�� x = -6,�� y = m x
+-1/sqrt(3)=+-sqrt(3)x/3
y=+-sqrt(3)x/3
x=-6,�� y =2sqrt(3)
P=(-6,2sqrt(3))
Can you please provide further clarification-�
Step 3 and 4-
m = 1/sqrt(3), x = -6, y = m x�
m =-1/sqrt(3), x = -6, y = m x�
How was the x coordinate found.�
Step 5�
+-1/sqrt(3)=+-sqrt(3)x/3�
what is happening here for one to become the other.
I need this explained very simply So i i can understand it. Especially my queries about the steps. Thank you very much.

Found 2 solutions by josgarithmetic, KMST:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
I just did this for you last night. Did you not understand my explanation?

Same as question 864931
http://www.algebra.com/algebra/homework/Linear-equations/Linear-equations.faq.question.864931.html
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
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:
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 , and the hypotenuse OC is .

That tells me that the angle measures
and OPC is a 30-60-90 triangle.
From the tangent of that angle you can also get the slopes of the tangents.
.
We multiply times because we do not like to see square roots in denominators,
so m = +- .
Of course, OP is the line with the negative slope, and OQ is the line with the positive slope.
For OP, and
for OQ, .
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 , the ratio is also the same for CPM, and OPM, so
and the x-coordinate of points M and P (and Q) is
For OP,
and for P, .
On the other hand, for Q, .

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 and center at is
.
For the circle in your problem, that would be
---> --->
If is tangent to the circle, there is only one point in common to circle and line,
so has one and only one solution.
-->-->-->-->
For what value of would that quadratic equation have one and only one solution?
A quadratic has one and only one real solution when ,
and then
In the case of ,
, , and .
It will have one and only one real solution when
-->-->-->-->
So -->-->-->
m = +-= +- = +- .
In either case, we can use to find the same value:
-->
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):

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