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How do you answer this?
Find the point(s) of intersection, if any, between each circle and line with the equation given:
1. (x-2)^{2} + (y+3)^{2} = 18
y = -2x - 2


Sagot :

To get the point of intersection, you must equate the two equations and solve for x and y.

[tex](x-2)^{2} + (y+3)^{2} = 18 -> equation1 \\ y=-2x-2 -> equation 2\\ x^{2}-4x+4 + (-2x-2+3)^{2} = 18 \\ x^{2}-4x+4 + (-2x+1)^{2} = 18 \\ x^{2}-4x+4 + 4x^{2} -4x+1=18\\ 5x^{2}-8x+5=18\\ 5x^{2}-8x-13=0\\ (5x-13)(x+1)=0\\ 5x-13 = 0 | x+1 = 0\\ x=\frac{13}{5} | x=-1\\ [/tex]

[tex]y=-2x-2 \\ y = -2(\frac{13}{5})-2 | y=-2(-1)-2 \\ y=\frac{-26}{5}-2 | y=2-2 \\ y=\frac{-36}{5}|y=0 \\ (\frac{13}{5},\frac{-36}{5}) and (-1,0)[/tex]