What is the equation of the circle with a center at #(6 ,3 )# and a radius of #2 #?

2 Answers
Jim G.
Mar 15, 2016

# (x - 6)^2 + (y - 3)^2 = 4#

Explanation:

The standard form of the equation of a circle is :

# (x - a )^2 + (y - b)^2 = r^2 #

where (a , b ) are the coords of the centre and r , the radius

here (a , b ) = (6 , 3 ) and r = 2

substitute these values into the standard equation

# rArr (x - 6 )^2 + (y - 3 )^2 = 4 " is the equation "#

George C.
Mar 15, 2016

#(x-6)^2+(y-3)^2=2^2#

or if you prefer:

#x^2+y^2-12x-6y+41 = 0#

Explanation:

The equation of a circle with centre #(h, k)# and radius #r# may be written:

#(x-h)^2+(y-k)^2 = r^2#

In our case #(h, k) = (6, 3)# and #r=2# so we get:

#(x-6)^2+(y-3)^2 = 2^2#

One nice thing about this form is that it is easy to read off the coordinates of the centre and the radius.

If you prefer, this can be put in standard polynomial form by multiplying out and rearranging a little:

#(x^2-12x+36)+(y^2-6y+9) = 4#

#->#

#x^2+y^2-12x-6y+41 = 0#

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