How do you write the equation of a circle with center(3,-4)and radius of 7?

1 Answer
Dec 11, 2015

Answer:

#x^2+y^2-6x+8y = 24#

Explanation:

The unit circle, centered in the origin, has equation #x^2+y^2=1#

This means that the squared lenght of the vector #(0.0)\to(x,y)# is one. So, a circle with radius #r#, centered in the origin, will have equation

#x^2+y^2=r^2#.

Now we want to move the center as well, let's say that our new center is #(x_0,y_0)#. To do so, we can create two additional variable #w# and #z# such that

#w=x-x_0#
#z=y-y_0#

With respect to these new coordinates, the circle is again centered in the origin, so it has equation

#w^2+z^2=r^2#

Plug back the definition, and you have the final result

#(x-x_0)^2 + (y-y_0)^2 = r^2#.

Plugging your particular values, the equation is

#(x-3)^2 + (y+4)^2 = 7^2#

If you like, you can do some manipulations:

#x^2-6x+9+y^2+8y+16=49#

#x^2+y^2-6x+8y = 24#