SOLUTION: Find the equation of the line (in slope-intercept form) that passes through the origin and the center of the circle 2x^2 + 2y^2 + 12x + 8y + 9 = 0.
Algebra ->
Graphs
-> SOLUTION: Find the equation of the line (in slope-intercept form) that passes through the origin and the center of the circle 2x^2 + 2y^2 + 12x + 8y + 9 = 0.
Log On
Question 54479: Find the equation of the line (in slope-intercept form) that passes through the origin and the center of the circle 2x^2 + 2y^2 + 12x + 8y + 9 = 0. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Find the equation of the line (in slope-intercept form) that passes through the origin and the center of the circle 2x^2 + 2y^2 + 12x + 8y + 9 = 0.
==========
Put the circle equation in the center-radius form, as follows:
2(x^2+x+?) + 2(y^2+4y+?)=-9
Complete the square to get:
2(x^2+x+1/4) + 2(y^2+4y+4)= -9+(1/2)+8
As you can see, this circle has a negative radius.
Might as well stop here.
Cheers,
Stan H.