# Intersection of a circle question! Finding where the circle pairs intersect.Find where the following circle pairs intersect. `x^2 + y^2 - 2x - 6y = 0` and `x^2 + y^2 - 4x - 2y = 0` In the...

Intersection of a circle question! Finding where the circle pairs intersect.

Find where the following circle pairs intersect.

`x^2 + y^2 - 2x - 6y = 0` and `x^2 + y^2 - 4x - 2y = 0`

In the example it shows how you make each equation in the form of `x^2 + y^2 =` ...... and equating each other, then finding y value. And they substituted y into the first equation to get x value, and then lastly substituting x into the y equatio to find y.

However I find this one hard because there are no numbers in the equation...

Please explain!

### 1 Answer | Add Yours

Let,

Circle1: `x^2+y^2-2x-6y = 0`

Circle2: `x^2 +y^2-4x-2y = 0`

To isolate `x^2+y^2 ` in each equations, move the other terms to the right side of the equation.

For Circle1,

`x^2+y^2-2x-6y = 0`

`x^2 + y^2 = 2x + 6y` ==> Let this be EQ.1 .

For Circle 2,

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

`x^2 + y^2 = 4x + 2y` ==> Let this be EQ.2 .

Then, set EQ.1 equal to EQ.2. And substitute.

`x^2 + y^2 = x^2 + y^2`

`2x + 6y = 4x+2y`

`6y - 2y = 4x - 2x`

`4y = 2x`

`2y = x`

` y = x/2`

Substitute `y=x/2` to either Circle1 or Circle2.

`x^2 + y^2 - 2x - 6y = 0`

`x^2 + (x/2)^2 - 2x - 6*(x/2) = 0`

`x^2 + x^2/4 - 2x - 3x = 0`

`x^2 + x^2/4 - 5x = 0`

To simplify, multiply both sides by 4.

`4(x^2 + x^2/4 - 5x) = 0*4`

`4x^2 + x^2 - 20x = 0 `

`5x^2 -20x = 0`

Then, factor.

`5x(x - 4) = 0`

Set each factor equal to zero and solve for x.

`5x = 0` and `x - 4 = 0`

`x = 0 ` `x = 4`

Substitute values of x to `y=x/2` . Determine value of y.

`x = 0` , `y = x/2 = 0/2 = 0`

`x = 4` , `y = x/2 = 4/2 = 2`

**Hence, the two circles intersect at points (0,0) and (4,2) . **