The equation of a circle is #x^2+(y+5)^2=81#. What is the center and radius of the circle?

2 Answers
May 16, 2017

#C(0,-5)#

#r=9#

Explanation:

The equation of a circle is given by #C_1=(x-a)^2+(y-b)^2=r^2#
where #(a,b)# is the centre and #r# is the radius.

We have #C_1=x^2+(y+5)^2=81#

We can rewrite this as #C_1=(x-0)^2+(y-(-5))^2=9^2#. So the centre of #C_1# is #(0,-5)# and the radius is #9#.

May 16, 2017

Centre#->(x,y)=(x,-5)#
Radius #->sqrt(81)=9#

Explanation:

Proper word for 'moving' is 'transpose'.

Suppose you had some function of #x# that we call #f(x)#

Then this will form a particular shape depending on what the form of #f(x)# is

If we changed this to #f(x-5)# then it will still be the same shape as #f(x)# but it would have 'shifted' right by 5.

On the other hand #f(x+5)# would have 'shifted' the graph of #f(x)# left by 5.

So for the x-axis:

a plus 'shifts' left
a minus 'shifts' right.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Similarly for y-axis:

a plus 'shifts' down
a minus 'shifts' up

Tony B