Question 456470
I'm pretty sure that trigonometry is needed for this problem, but I'm not totally sure!

The line 3y+x=10 intersects the circle x^2+y^2=50 in two distinct points. Suppose you drew a radius to each of the two points of intersection. Find the angle, in standard position, created by each radius.

Thank you!!!!
<pre>
The other tutor's second angle is incorrect.  It should be 135°.
He thought you wanted the angle between the two radii, not the
angles in standard position.

{{{drawing(600,600,-10,10,-10,10, graph(600,600,-10,10,-10,10),
green(line(0,0,-5,5), line(0,0,7,1)),
circle(0,0,sqrt(50)), line(-11,7,13,-1) )}}}

First we'll find the two points where the line intersects the
circle:

{{{system(3y+x=10,x^2+y^2=50)}}}

Solve the first equation for x

{{{x=10-3y}}}

Substitute in the second equation:

{{{x^2+y^2=50}}}

{{{(10-3y)^2+y^2=50}}}
{{{100-60y+9y^2+y^2=50}}}
{{{100-60y+10y^2=50}}}
{{{10y^2-60y+50=0}}}
Divide through by 10
{{{y^2-6y+5=0}}}
Factor
{{{(y-1)(y-5)=0}}}
Use zero factor property:
{{{y-1=0}}} so {{{y=1}}}
{{{y-5=0}}} so {{{y=5}}}

To find the x-value correponding to each of those,
substitute each y-value into {{{x=10-3y}}}

To find the x-value that corresponds to y=1 
{{{x=10-3y}}}
{{{x=10-3(1)}}}
{{{x=10-3}}}
{{{x=7}}}

Therefore one point of intersecion is (7,1).

To find the x-value that corresponds to y=5 
{{{x=10-3y}}}
{{{x=10-3(5)}}}
{{{x=10-15}}}
{{{x=-5}}}

Therefore the other point of intersection is (5,-5).

{{{drawing(600,600,-10,10,-10,10, graph(600,600,-10,10,-10,10),
green(line(0,0,-5,5), line(0,0,7,1)), locate(-7,5,"(-5,5)"),locate(7,1.7,"(7,1)"),
circle(0,0,sqrt(50)), line(-11,7,13,-1) )}}}

We'll take each one separately:
Let's erase everything except the radius on the right:

{{{drawing(600,600,-10,10,-10,10, graph(600,600,-10,10,-10,10),
green( line(0,0,7,1)),locate(7,1.7,"(7,1)") )}}}

Let's draw a perpendicular from the point (7,1) to the x-axis,
and label the angle &#5054;, indicated by the red counterclockwise arc.

{{{drawing(600,600,-10,10,-10,10, graph(600,600,-10,10,-10,10),
green( line(0,0,7,1)),locate(7,1.7,"(7,1)"), line(7,1,7,0),
red(arc(0,0,9,-9,0,8.130102354)), locate(5,.6,theta)

 )}}}

{{{tan(theta) = y/x}}}
{{{tan(theta) = 1/7}}}
{{{theta = "8.130102354°"}}}

That's one answer.   

Now let's erase everything except the radius on the leftt:

{{{drawing(600,600,-10,10,-10,10, graph(600,600,-10,10,-10,10),
green( line(0,0,-5,5)),locate(-7,5,"(-5,5)") )}}}

Let's draw a perpendicular from the point (7,1) to the x-axis,
and label the angle &#5054;, indicated by the red counterclockwise arc.

{{{drawing(600,600,-10,10,-10,10, graph(600,600,-10,10,-10,10),
green(line(0,0,-5,5)),locate(-7,5,"(-5,5)"), line(-5,5,-5,0),
red(arc(0,0,9,-9,0,135)), locate(3,4,theta), locate(-1.7,.7,"45°"),
blue(arc(0,0,4,-4,135,180))
 )}}}

{{{tan(theta) = y/x}}}
{{{tan(theta) = -5/5}}}
{{{tan(theta) = -1}}}
So the reference angle for &#5054; is 45° degrees.
And it's in the 2nd quadrant so we subtract from 180°
and get 180°-45° = 135°

So the second angle is

{{{theta = "135°"}}}

Edwin</pre>