well, y = 5-x, so
x^2 + (5-x)^2 = 25
now just expand and solve the quadratic.
Find the points of intersection of the line X+Y=5, and the circle x^2+y^2=25 by solving the system of equations.
4 answers
x+5/30=4x-10/75
2*3*5*5*(x+5)/2*3*5=2*3*5*5(4x-10)/3*5*5
5*(x+5)=2*(4x-10)
5x+25=8x-20
25+20=3x
3x=45
x=15
2*3*5*5*(x+5)/2*3*5=2*3*5*5(4x-10)/3*5*5
5*(x+5)=2*(4x-10)
5x+25=8x-20
25+20=3x
3x=45
x=15
x=0, y=0
x=5,y=25
no other solutions
x=5,y=25
no other solutions
The x=15 is ok
Not sure what the last set of values is for, but if they are for the 1st problem, no good.
x^2 + (5-x)^2 = 25
x^2 + x^2-10x+25 = 25
2x^2-10x = 0
2x(x-5) = 0
x = 0 or x=5
So, the points of intersection are (0,5) and (5,0)
Not sure what the last set of values is for, but if they are for the 1st problem, no good.
x^2 + (5-x)^2 = 25
x^2 + x^2-10x+25 = 25
2x^2-10x = 0
2x(x-5) = 0
x = 0 or x=5
So, the points of intersection are (0,5) and (5,0)