189k views
3 votes
Rewrite the equation of the circle (x-0.8)^2+(y+0.6)^2=0.2 in general form

1 Answer

5 votes
Answer: x^2+y^2-1.6x+1.2y+0.8=0

--------------------------------------------------------------------

Work Shown:

Expand out the x expression first
(x-0.8)^2 = (x-0.8)(x-0.8)
(x-0.8)^2 = x(x-0.8)-0.8(x-0.8)
(x-0.8)^2 = x^2-0.8x-0.8x+0.64
(x-0.8)^2 = x^2-1.6x+0.64

Then do the y
(y+0.6)^2 = (y+0.6)(y+0.6)
(y+0.6)^2 = y(y+0.6)+0.6(y+0.6)
(y+0.6)^2 = y^2+0.6y+0.6y+0.36
(y+0.6)^2 = y^2+1.2y+0.36

So overall we can say the following
(x-0.8)^2+(y+0.6)^2=0.2
x^2-1.6x+0.64+(y+0.6)^2=0.2
x^2-1.6x+0.64+y^2+1.2y+0.36=0.2
x^2+y^2-1.6x+1.2y+0.64+0.36=0.2
x^2+y^2-1.6x+1.2y+0.64+0.36-0.2=0
x^2+y^2-1.6x+1.2y+0.8=0
which is the final answer

User StelioK
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.