The 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)
Solve the system
x^2+y^2-4x+2y=20
4x +3y =5
1 answer