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)).`
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