How do you write the standard form of the following equation then graph x² + y² - 4x + 14y - 47 = 0 ?
1 Answer
Mar 20, 2018
We do a double completion of square--one with the
#1(x^2 - 4x + 4- 4) + 1(y^2 + 14y + 49 - 49) - 47 = 0#
#1(x^2 - 4x + 4) - 4 + 1(y^2 + 14y + 49) - 49 - 47 = 0#
#(x - 2)^2 - 4 + (y + 7)^2 - 49 - 47 = 0#
#(x -2)^2 + (y + 7)^2 = 100#
Now recall that when a circle is in the form
Therefore this is a circle with centre
Hopefully this helps!