Question 1169814

A circle is centered at the origin and has a radius of √5 units. A line with a slope of 2 passes
through the origin and intersects the circle in two places. Where does the line intersect the
circle?

This is the last question that Im struggling please help me
<pre>*[illustration ADC_1169814_3.png].
m (slope) = 2  ; Point (0, 0) 
Equation of line that passes through origin (also center of circle): y = 2x
Equation of circle:	{{{matrix(3,3, (x  -  h)^2 + (y  -  k)^2, "=", r^2, 			(x  -  0)^2 + (y  -  0)^2, "=", (sqrt(5))^2, x^2 + y^2, "=", 5)}}}
We now have the following system of equations:	{{{matrix(2,6, y, "=", 2x, "--------", eq, "(i)", x^2 + y^2, "=", 5, "----", eq, "(ii)")}}}
                                                {{{matrix(1,3, x^2 + (2x)^2, "=", 5)}}} ------- Substituting 2x for y in eq (ii)
                                                {{{matrix(5,3, x^2 + 4x^2, "=", 5, 5x^2, "=", 5, 5(x^2), "=", 5(1), x^2, "=", 1, x, "=", " "+- 1)}}}
x = 1						x = - 1
y = 2x						y = 2x
y = 2(1) = 2					y = 2(- 1) = - 2
Intersection: (1, 2)				Intersection: (- 1, - 2)
Therefore, the line intersects the circle at {{{highlight_green(matrix(1,5, "(1,", "2)", and, "(- 1,", "- 2)"))}}} (see above diagram).
NOTE that the equation of the line: y = 2x "states" that the value of the y-coordinate is TWICE that of the x=coordinate.