# Determine the solution set for the following (x-2)^2 + (y-3)^2 = 4 and x+y = 4

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`(x-2)^2+(y-3)^2 = 4`

`x+y =4`

Let's substitute for y from second equation to equation1,

`y = 4-x`

Then,

`(x-2)^2+(4-x-3)^2 = 4`

`(x-2)^2+(-x+1)^2 = 4`

`(x-2)^2+(x-1)^2 = 4`

`x^2-4x+4+x^2-2x+1=4`

`2x^2-6x+1 = 0`

`x = (6+-sqrt((-6)^2-4(2)(1)))/(2*2)`

`x = (6+-sqrt(36-8))/4`

`x = (3+-sqrt(7))/2`

Therefore solutions for x are,

`x = (3+sqrt(7))/2`

and

`x = (3-sqrt(7))/2`

The respective solutions for y are,

`x = (3+sqrt(7))/2`

`y = 4 - (3+sqrt(7))/2 `

`y = (5-sqrt(7))/2 `

AND,

`x = (3-sqrt(7))/2`

`y = 4- (3-sqrt(7))/2`

`y = (5+sqrt(7))/2`

So, this is a line and a circle. It could have two solutions, or one, or none.

One way to do this is to substitute; solve the second equation for x, for example:

x = 4-y

Now substitute:

(4-y-2)^2+(y-3)^2=4

(2-y)^2+(y-3)^2=4

(4-4y+y^2)+(y^2-6y+9)=4

4-4y+y^2+y^2-6y+9=4

2y^2-10y=9=0

This is not factorable, so use the quadratic formula. This gives two solutions:

y=(10+-sqrt(28))/4

Subtract each of those from 4 to get the x values.