SOLUTION: Find the equation of the diameter of the circle x^2+y^2-4x+6y=14 which passes through the origin

Algebra ->  Linear-equations -> SOLUTION: Find the equation of the diameter of the circle x^2+y^2-4x+6y=14 which passes through the origin      Log On


   

Question 941342: Find the equation of the diameter of the circle x^2+y^2-4x+6y=14 which passes through the origin
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-4x%2By%5E2%2B6y=14
x%5E2-4x%2B4%2By%5E2%2B6y%2B9=14%2B4%2B9
%28x-2%29%5E2%2B%28y%2B3%29%5E2=27

The LINE containing the diameter would contain the circle's center of (2,-3), AND also the origin (0,0). You could put limitations on the domain for the line to achieve just the diameter segment; but in any case, you can find the line containing those two points. The equation for the line is y=-%283%2F2%29x.

Find the endpoints of the diameter using the equation of the circle and the equation y=-(3/2)x. Substitute for y in the circle equation, solve for x, and find the corresponding y value.... you have the endpoints of the diameter.