(x+2)^2+y^2=4 y-4=x

Question

(x+2)^2+y^2=4 y-4=x

Answer 1

X^2+2*x*2+2^2+y^2=4      y=x+4
x^2+4x+4+(x+4)^2=4
x^2+4x+4+x^2+2*x*4+4^2=4
2x^2+12x+16=0
x^2+6x+8=0
D=6^2-4*1*8=36-32=4
X1=(-6+корень(4))/2*1= (-6+2)/2=-4/2=-2
X2= (-6-корень(4))/2*1= (-6-2)/2=-8/2=-4

Comments

ну и Y1= -2+4=2 Y2= -4+4=0