Solve the system:

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

x-y^2 = -4

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Let `(x-1)^2 + y^2 = 25` be equation (i)

`x-y^2 = -4` .........equation (ii)

From (ii) we get:

`y^2=x+4`

Substituting the value of `y^2` in eq. (i) we get:

`(x-1)^2 + x+4 = 25`

`rArr x^2-2x+1+x+4=25`

`rArr x^2-x-20=0`

Now factorize by splitting the middle term,

`x^2-5x+4x-20=0`

`rArr x(x-5)+4(x-5)=0`

`rArr (x-5)(x+4)=0`

Using zero product property:

For (x-5)=0 we get x=5

For (x+4)=0 we get x=-4

Now substitute each value of x in eq.(ii) to obtain the corresponding values of y:

For x=5, `y^2=5+4=9` ` rArr y=+-3`

For x=-4 `y^2=-4+4=0` ` rArr y=0`

**Therefore, the solution set is `{(5,+-3),(-4,0)}` .**

**Sources:**

Isolate and substitute.

Isolate y^2, so you get y^2 = x+4

Substitute y^2 into the other equation, so you get: (x-1)^2 + x + 4 = 25

Simplify:

x^2 - x - x + 1 + x + 4 = 25

x^2 - x + 5 = 25

x^2 - x - 20 = 0

Factor: (x-5) (x+4) = 0

x-5=0, so x=5

x+4=0, so x= -4

Solutions: x=5, x=-4

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