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waheguru
Group Title
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?
 one year ago
 one year ago
waheguru Group Title
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?
 one year ago
 one year ago

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dake Group TitleBest ResponseYou've already chosen the best response.0
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=103y x^2+y^2=50 (103y)^2+y^2=50 10060y+9y^2+y^2=50 10060y+10y^2=50 divide by 10 106y+y^2=5 y^26y+5=0 (y5)(y1)=0 y=5 x=103y=1015=5 y=1 x=103y=103=7 .. Points of intersection: (5,5) and (7,1) .. y=(50x^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 1808.1345=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.
 one year ago

waheguru Group TitleBest ResponseYou've already chosen the best response.0
This is a very complicated way. We learned a very simple way but I canno't remember
 one year ago

dake Group TitleBest ResponseYou've already chosen the best response.0
give me a medal
 one year ago

amir.sat Group TitleBest ResponseYou've already chosen the best response.1
\[x ^{2}+y ^{2}=50\] so bring the y in one side \[y=\pm \sqrt{50x ^{2}}\] now whatever number instead of x and get it's y
 one year ago

waheguru Group TitleBest ResponseYou've already chosen the best response.0
Thank you
 one year ago
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