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

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thilina-g's profile pic

Posted on

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

`x+y =4`


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

`y = 4-x`


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

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

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


`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 `


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

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

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


baronridiculous's profile pic

Posted on

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:






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


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

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