Findind the intersection of two circles.?

Question

Could someone help me find the intersections of these pairs of equations (and write out the steps neccessary to find them -

x^2 + y^2 = 4, x^2 + (y-6)^2 = 25
x^2 + y^2 - 4y = 0, x^2 + y^2 - 2x = 4
x^2 + y^2 = 4, x^2 + y^2 - 6x - 6y = -14

Cheers!

Cooper

Answer 1

Easiest way is to write them as two equations equal to zero. Then equate them...

Starting with your first example:
x² + y² - 4 = 0
x² + (y-6)² - 25 = 0

Now equate these:
x² + y² - 4 = x² + (y-6)² - 25

Expand out the (y-6)²
x² + y² - 4 = x² + (y²-12y+36) - 25

Now cancel like terms (like x², y², etc.) which are on both sides:
-4 = -12y + 36 - 25

Now simplify:
12y = 36 - 25 + 4
12y = 15
y = 15/12
y = 5/4

Now use the value you found (y=5/4) and solve for the other variable using one of the original equations:
x² + (5/4)² = 4
x² + (25/16) = 4
x² = 64/16 - 25/16
x² = 39/16
Here you need to take the sqrt and you'll have two answers:
x = sqrt(39/16)
x = +/- sqrt(39)/4

So your two points of intersection for the first pair of equations are:
(sqrt(39)/4, 5/4) and (-sqrt(39)/4, 5/4)

The same method works for the other problems...

Answer 2

wow, i dont have an answer i just wanted to say that is the most confusing thing ive ever seen in my life!!