Centre/radius of Circles (1 Viewer)

[holdenkicks]

I got a maths poblem. Studying for yr 12 Half yearlies, as Im sure many of you are aswell.

question:
Find the centre and radius of the circle with equation
x^2 + y^2 + 8x - 2y - 8 = 0


My teacher told me I had half it and square it, thus i got:

x^2 + 8x + 16 + y^2 - 2y +1 = 16+1+8

But im not sure what to do after this. Any help with working would be greatly appreciated. thx.

 

[jesshika]

okai once you've done that...
yu factorise the eqnzz

..... so therefore it bocomes
(x+4)^2 + (y-1)^2 = 25

so now you can see that the centre of the circle is (-4, 1) with rradius 5 ...

you do know why right??
x^2 + y^2 = r^2
with centre at the origin and radius r

 

[holdenkicks]

thx for that.

however I dont understand how u got taht when u differentiated.

u sed:
(x+4)^2 + (y-1)^2 = 25

but i get
2x + 8 + 2y - 2 when i differentiated

can u explain what u did to get (x+4)^2 + (y-1)^2 = 25?

 

[martin310015]

Originally posted by holdenkicks
\
question:
Find the centre and radius of the circle with equation
x^2 + y^2 + 8x - 2y - 8 = 0

for this question u don't need to differentiate...just complete the square.

1. x^2+y^2+8x-2y-8=0

2.x^2+8x+y^2-2y=8

3.(x^2+8x+16)+(y^2-2y+1)=8+16+1......this part is completing the square.

4 (x+4)^2+(y-1)^2=25..........you factorise now

5. so has jesshika said it is a circle with centre (-4,1) and radius 5 units

hope that helped

 

[Collin]

Like the above has stated, differentiation is irrelevant here.

You simply need to complete the square on the equation in order to obtain the form (x-h)^2 + (y-k)^2 = r^2, of Centre (h,k) and Radius r.