Question 203604
Find the point of the intersection of the line y=x+2 and the circle 
(x-1)^2 + y^2=16
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Substitute for "y" to get:
(x-1)^2 + (x+2)^2 = 16
x^2-2x+1 + x^2+4x+4 = 16
2x^2 +2x + 5 = 16
2x^2 + 2x - 9 = 0
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x = [-2 +- sqrt(4 - 4*2*-9)]/4
x = [-2 +- sqrt(4 + 72)]/4
x = [-2 +- 2sqrt(19)]/4
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x = (-1/2) + (1/2)sqrt(19) or x = (-1/2)-(1/2)sqrt(19)
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Substitute each of these into y = x+2 to get the y-values corresponding to the
x values:
Those y values will be (3/2)+(1/2)sqrt(19) and (3/2)-(1/2)sqrt(19)
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Cheers,
Stan H.
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