Not to difficult, all we have to do is think about it.

What info do we have?

Well we know that x + y = -1

now that means that y = -1 - x

sub this in to out x^2 + y ^2 =25 formula and we get

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

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

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

x^2 + x^2 +2x +1 -25 = 0

2x^2 + 2x -24 = 0

x^2 + x -12 = 0

By the cross method (x+4)(x-3) = 0

x = -4 and x = 3

Alternatively you can use quadratic fomula:

x = (-b +- ((b^2 - 4ac)^1/2))/2a

x = (-1 +- ( 1 + 48)^1/2)/2

x = (-1 +- 7)/2

x= -4 and 3

Therefore subing in these x values into x + y = -1:

-4 + y = -1

y = 3

Therefore one point is (-4,3)

3 + y = -1

y = -4

Therefore one point is (3,-4)

Therefore the 2 functions (actually i should say curves given that a circle isn't a function) meet at (-4,3) and (3, -4)

Edit: Could someone please tell me where I can find the code to put in symbols and equations, such as square root signs instead of ^1/2 , integral signs and dy/dx etc. Much appreciated