SOLUTION: on the circle x^2+y^2+2x-8y+7=0. Find points at distance 5 from (2,-2)

Question 631982: on the circle x^2+y^2+2x-8y+7=0. Find points at distance 5 from (2,-2)
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
on the circle x^2+y^2+2x-8y+7=0. Find points at distance 5 from (2,-2)
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The 2 points are the intersection of the given circle and a circle of radius 5 about the given point.
The 2nd circle is (x-2)^2 + (y+2)^2 = 25
x^2 - 4x + 4 + y^2 + 4y + 4 = 25
x^2 + y^2 - 4x + 4y = 17
x^2 + y^2 + 2x - 8y = -7
--------------------------- Subtract
-6x + 12y = 24
x = 2y -4 is the eqn of the line thru the 2 points
Sub for x in either equation of a circle.
x^2 + y^2 - 4x + 4y = 17
(2y-4)^2 + y^2 - 4(2y-4) + 4y = 17
5y^2 - 20y + 15 = 0
y^2 - 4y + 3 = 0
y = 1, x = -2 --> (-2,1)
-------
y = 3, x = 2 --> (2,3)

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