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waheguru
I am confused on how to find points on a circle. I am given that the equation is x^2+y^2=50 From here how do I find two points?
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. .. 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 .. Points of intersection: (-5,5) and (7,1) .. 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.
This is a very complicated way. We learned a very simple way but I canno't remember
\[x ^{2}+y ^{2}=50\] so bring the y in one side \[y=\pm \sqrt{50-x ^{2}}\] now whatever number instead of x and get it's y