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...
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.
`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`
`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) .