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Find all points (x,y) that are 12 units away from the point (3,-8) and that lie on the line y = 2x + 1.

Dec 8, 2020

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The intersection points will be  where  the line with the equation  ( x - 3)^2 + ( y + 8)^2 = 12^2   intersects  the line  y  = 2x +1

Sub  the equation of the line into the equation of the  circle  for  y  and we have

(x - 3)^2  +  [ (2x + 1) + 8 ]^2  =  12^2      simplify

x^2  - 6x  +  9  +   ( 2x + 9)^2   =144

x^2  - 6x + 9  +  4x^2  + 36x  +  81   = 144

5x^2  + 30x  + 90  = 144        subtract 144 from both sides

5x^2  + 30x  - 54  =  0

x =   -30  +  sqrt  ( 900 + 20 *54)              -30 + sqrt (1980)            -30 + 6sqrt (55)

______________________  =       ________________  =  ________________  =

2 * 5                                                   10                                 10

[-15 + 3sqrt ( 55)]  / 5

And the other  x value  is    [-15  -3sqrt (55) ] / 5

So  y =    (2/5) ( -15  + 3sqrt (55)) + 1  =   -6  + (6/5)sqrt (55)  +1  =   -5 + (6/5)sqrt (55)

And  the other y  =  -5 - (6/5)sqrt (55)

So  the intersection points are

(  [ -15 + 3sqrt (55)] / 5, -5 + (6/5)sqrt (55)  )    and  ( [ -15 -3sqrt (55) ] / 5  , -5  - (6/5)sqrt (55) ]   Dec 8, 2020