When graphed, the circle with equation
x2 + y2 - 14x + 10y + 65 = 0
will lie entirely in quadrant

Hello,

First we must use the property of the sum of perfect squares:

X^2 -14x +y^2 +10y+65 =0

Adding 7^2 and -7^2

x^2 -14x+7^2-7^2 +y^2+10y+65=0

x^2 -2.7x +7^2 -49 +y^2+10y+65=0

(X-7)^2 +y^2+10y+16=0

Adding 5^2 and -5^2

(X-7)^2 +y^2 +10y+5^2-5^2+16=0

(X-7)^2 +y^2+2.5y+5^2 -25 +16=0

(X-7)^2+ (y+5)^2 -9 = 0

(X-7)^2 +(y+5)^2 = 9

(X-7)^2+(y+5)^2 = 3^2

Center = (7, -5)
R = 3

The circle is in the 4 quadrant.

Gotta be d the music of previous eras