Question

Find the intersection of the two circles x²+y²-2x+4y-11=0 and x²+y²+4x+2y-9=0.

Answer 1

(0.3135, 1.9405) and (-1.9135, -4.7405)

Explanation:

`x^2+y^2-2x+4y-11=0`

`x^2+y^2+4x+2y-9=0`

Subtracting the equations we get

`-6x+2y-2=0\\\Rightarrow -3x+y-1=0\\\Rightarrow y=1+3x`

Putting the y value in the first equation

`x^2+(1+3x)^2-2x+4(1+3x)-11=0\\\Rightarrow x^2+1+9x^2+6x-2x+4+12x-11=0\\\Rightarrow 10x^2+16x-6=0\\\Rightarrow 5x^2+8x-3=0`

`x=(-8+√(8^2-4\cdot \:5\left(-3\right)))/(2\cdot \:5), x=(-8-√(8^2-4\cdot \:5\left(-3\right)))/(2\cdot \:5)\\\Rightarrow x=0.3135, -1.9135`

`y=1+3x\\\Rightarrow y=1+3(0.3135)\\\Rightarrow y=1.9405`

`y=1+3x\\\Rightarrow y=1+3(-1.9135)\\\Rightarrow y=-4.7405`

So, the points of intersection are (0.3135, 1.9405) and (-1.9135, -4.7405)