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September 30, 2014

September 30, 2014

Posted by **Jill** on Monday, March 12, 2012 at 3:30pm.

x^2+y^2-4x+2y=20

4x +3y =5

- math -
**MathMate**, Monday, March 12, 2012 at 3:39pmThe first one is a circle centred at (2,-1), or

(x-2)^2+(y+1)^2=20+4+1=25

or

C: (x-2)^2+(y+1)^2=5^2

The second equation is a straight line.

L: 4x+3y=5

To solve the system, set

y=(5-4x)/3

and substitute into equation C

or

(x-2)^2+((5-4x)/3+1)^2=25

Solve for x to get x=5 or x=-1

Substitute into y=(5-4x)/3

to get

y=-5, y=3

So the intersections are

(5,-5), or (-1,3)

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