Point of Intersection between 2 circles?

Help me out please, ive been searching around and cant find any understandable solutions..

1. Find the coordinates of the points of intersection of each of the following pairs of circles :
x^2 + y^2 - 3x + 5y - 4 = 0 and
x^2 + y^2 - x + 4y - 7 = 0

2. Show that the circles x^2 + y^2 - 10x - 8y + 18 = 0 and x^2 + y^2 - 8x - 4y + 14 = 0 do not intersect each other

3. Two circles C1, C2 have equations (x+1)^2 + y^2 = 9 and x^2 + y^2 - 16x - 80 = 0 respectively
i) find radius of C1 and C2
ii) calculate the distance between the centers of C1 and C2
iii) show that C1 and C2 touch each other and find the coordinates of T, their point of contact

Please show step by step working cause i am afraid i might get confused. Thank you

You have the same leading coefficients in the two equations, so If you take their difference, the square terms will go away.

x² + y² - x + 4y - 7 = 0
x² + y² - 3x + 5y - 4 = 0
-----------------------------------
2x - y - 3 = 0

This line is the radical axis of the two circles. If they intersect, then it is their common chord produced. If the circles do not intersect, then the radical axis will not intersect either of them.

Solve for y, and substitute it into either of the circle equations.

y = 2x - 3
x² + (2x - 3)² - x + 4(2x - 3) - 7 = 0
5x² - 5x - 10 = 0
x² - x - 2 = 0
(x + 1)(x - 2) = 0
x = -1 or x = 2

Substitute these values intor the linear equation to find the corresponding y-coordinate.

Let x = -1
y = 2(-1) - 3 = -5
Solution: (-1, -5)

Let x = 2.
y = 2(2) - 3 = 1
Solution: (2, 1)

Those are the two points of intersection. If there were no circle intersection, the quadratic equation would have no real solutions. If the circles were tangent, the quadratic equation would have only one real root.