The circle intersects the line with he equation x+y=3 at 2 points, A and B. Find algebraically the coordinates A and B show the distance AB= sqrt 162 thanks :)

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The circle intersects the line with he equation x+y=3 at 2 points, A and B. Find algebraically the coordinates A and B show the distance AB= sqrt 162 thanks :)

Mathematics
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what line?
a circle equation has an equation of x6@+y^2=45 is that what you mean?
MissMys, you just wrote something in the comments? is that more info about this problem?

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yes :)
Well you have to write the whole question word for word as i was given to you, in order for us to be of any help.
a circle equation has an equation of x^2+y^2=45 The circle intersects the line with he equation x+y=3 at 2 points, A and B. Find algebraically the coordinates A and B show that the distance AB is sqrt 162 and the previous qs asks about radius and centre of the circle
Put the two equations together, you would find to values for y=3, y=34. You can use these values to obtain related values for x. The two x,y values should be point A and B. You can double check it with distance formula.
sorry but i dont understand? :S so do i put those two values into the equation??
You have your equation for the line, you can solve for x, or y. From your line equation, x=3-y. Put this in the circle equation. (3-y)^2+y^2=45. I have already done this much for you, the result is y=3, y=34
urm, i havent got a clue what you did and 34 is wrong :S i am so dead tomorrow :(
i understand the first bit then confused of how you got 3 and 34 :S
Well forget about 3 and 34. Solve for y from the new equation created.
thanks for helping but i cant do it :( ill still give u the medal :)
What can't you do? You have an equation (3-y)^2+y^2=45, solve for y.
yea i dunno what or how you do the squared bit :(
\[(3-y)^{2}=(3-y)(3-y)\]
thanks :)

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