How do you find the equation of the circle with radius 6 and center (2,4)?

2 Answers
Apr 16, 2016

I found: #x^2+y^2-4x-8y-16=0#

Explanation:

You can use the general relationship for the equation of a circle of center at #(h,k)# and radius #r# as:

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

In your case:
#(x-2)^2+(y-4)^2=6^2#
#x^2-4x+4+y^2-8y+16=36#
#x^2+y^2-4x-8y-16=0#

Apr 16, 2016

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

Explanation:

The standard form of the equation of a circle is.

#color(red)(|bar(ul(color(white)(a/a)color(black)( (x - a )^2 + (y - b)^2 = r^2)color(white)(a/a)|)))#
where (a , b) are the coordinates of the centre and r , the radius

in this question a = 2 , b = 4 and r = 6

substitute these values into the standard equation

#rArr (x - 2)^2 + (y - 4)^2 = 36" is the equation " #