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.
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Finding points of intersection
3y+x=10
x=10-3y
x^2+y^2=50
(10-3y)^2+y^2=50
100-60y+9y^2+y^2=50
100-60y+10y^2=50
divide by 10
10-6y+y^2=5
y^2-6y+5=0
(y-5)(y-1)=0
y=5
x=10-3y=10-15=-5
y=1
x=10-3y=10-3=7
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Points of intersection: (-5,5) and (7,1)
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y=(50-x^2)^.5
y=-x/3+10/3
if you draw a line joining the center with the points of intersection, you will get two reference angles.
The tangent of the reference angle on the right is 1/7. Taking the inverse tan of 1/7=8.13º. At the other point, the inverse tan=5/5=1=45º. The angle in standard position is 180-8.13-45=126.87º
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.