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

Expert Answers

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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`

 

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