# In how many points do the graphs of the equations x^2 + y^2 = 25 and y^2 = 4x intersect?

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You need to remember that the coordinates of points that lie on both curves need to be solutions of both given equations.

You need to express `y^2` in terms of `x^2` in the equation `x^2 + y^2 = 25 ` such that:

`y^2 = 25 - x^2`

You need to set the equations `y^2 = 25 - x^2` and `y^2 = 4x ` equal such that:

`25-x^2 = 4x`

`-x^2 - 4x + 25 = 0`

`x^2 + 4x - 25 = 0`

You should complete the square such that:

`x^2 + 4x + 4 - 4 - 25 = 0`

`(x+2)^2 - 29 = 0`

`(x+2)^2 = 29 =gt x+2 = +-sqrt29`

`x_1 = -2 + sqrt29`

`x_2 = -2 - sqrt29`

You need to find the coordinates `y_1` and `y_2` such that:

`y_(1,2) = +-sqrt(4(sqrt29-2))`

`y_(3,4) = +-sqrt(4(-sqrt29-2))`

**Hence, evaluating the coordinates of points of intersection of curves yields `x_1 = -2 + sqrt29 ; y_(1,2) = +-sqrt(4(sqrt29-2))` and `x_2 = -2 - sqrt29 ; y_(3,4) = +-sqrt(4(-sqrt29-2)).` **

We'll solve the system of the given equations, to verify how many solutions the system does have.

We'll substitute y^2 by 4x in the 1st equation:

x^2 + 4x - 25 = 0

We'll determine the solutions of the quadratic:

x1 = [-4+sqrt(16+100)]/2

x1 = (-4+2sqrt29)/2x1 = -2+sqrt29

x2 = -2-sqrt29

y^2 = 4*x1 =-8 + 4sqrt29

Since -8 + 4sqrt29 > 0 => y1 = sqrt(-8 + 4sqrt29) and y2 =

-sqrt(-8 + 4sqrt29)

y^2 = 4*x2 = -8 - 4sqrt29 < 0 => there are no real values for

y^2 = sqrt(-8 - 4sqrt29)

**The equations x^2 + y^2 = 25 and y^2 = 4x are intercepting in two points (-2+sqrt29 ; sqrt(-8 + 4sqrt29)) and (-2+sqrt29 ; -sqrt(-8 + 4sqrt29)). **