How do you find the equation of a circle passes through the origin and has its center at (-1,-9)?

1 Answer
Alan P.
Jun 20, 2016

#(x+1)^2+(y+9)^2=82#

Explanation:

If a circle has a center at #(x_c,y_c)=(-1,-9)# and passes through the origin (i.e. #(0,0)# )
then it has a radius of r=#sqrt((-1-0)^2+(-9-0)^2)=sqrt(82)#

the general formula for a circle with center #(x_c,y_c)# and radius #r# is
#color(white)("XXX")(x-x_c)^2+(y-y_c)^2=r^2#

So in this particular case:
#color(white)("XXX")(x-(-1))^2+(y-(-9))^2=82#
or
#color(white)("XXX")(x+1)^2+(y+9)^2=82#

graph{(x+1)^2+(y+9)^2=82 [-24.15, 21.45, -21.76, 1.06]}

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