How do you find the equation of a circle with center at the origin and passing through (-6,-2)?

1 Answer
Morgan
Mar 2, 2017

#x^2+y^2=40#

See below.

Explanation:

The equation of a circle with center #(h,k)# and radius #r# is given by #(x−h)^2+(y−k)^2=r^2#. For a circle centered at the origin, this becomes the more familiar equation #x^2+y^2=r^2#.

Because we know the circle is centered at the origin, i.e. #(0,0)#, we can use this fact along with the point which the circle passes through #(-6,-2)# to find the radius. This is done using the distance formula.

#r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

Given the points #(0,0)# and #(-6,-2)#:

#r=sqrt((-6-0)^2+(-2-0)^2)#

#=sqrt(36+4)#

#sqrt(40)=2sqrt(10)#

#=>r=2sqrt(10)#.

Therefore, the equation of the circle is given by:

#x^2+y^2=(2sqrt(10))^2#

#=>x^2+y^2=40#

graph{x^2+y^2=40 [-20, 20, -10, 10]}

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