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 w

Question 864931: 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
i do not understand what is happening, how was the x coordinate procured.
Step 5
+-1/sqrt(3)=+-sqrt(3)x/3
what is happening here for one to become the other.
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Line y=mx contains the origin as one of its points. We could guess that the intent of the problem is to find the lines tangent to the circle and containing the point (0,0). The description really needs to be exact, but the included origin point might be guessed as the intent.

The circle has two branches of functions. The upper branch is . A general point (x,y) on the upper branch of the circle is the variable ordered pair, (x, sqrt(16-(x+8)^2) ).

Center point of the circle is (-8,0).

The question needs two slopes which must be negative reciprocals.
Slope, Origin to tangency point on circle:
Slope, Center point to tangency point on circle:

Reminder, these slopes must be negative reciprocals:
;
Solve this for x.
Use the formula for the circle already solved for y, to get the corresponding value of y.
Now you have one of the (x,y) points for the tangency point, from which you can now compute the slope from (0,0) to this point (x,y) on the circle.
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