When graphed, what quadrant will the circle #x^ { 2} + y ^ { 2} - 14x + 10y + 65= 0# lie in?

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Answer 1 · 2017-05-10T05:30:32

#4th# quadrant

Given: #x^2 + y^2 - 14x + 10y + 65 = 0#

Use completing of the square to get the circle in the standard form

#(x-h)^2 + (y-k)^2 = r^2#, where the center #(h. k)# and #r# is the radius

#(x^2 -14x) + (y^2 + 10y) = - 65#

Half each #x#-term and add the square of this value to the right side of the equation:

#(x - 7)^2 + (y + 5)^2 = -65 + 49 + 25#

#(x - 7)^2 + (y + 5)^2 = 9#

Circle with center #(7, -5)# and #r = 3# will be found in the #4th# quadrant.