How do you find the general form of the equation of the circle with center (3,3) and containing the point (4,4)?

1 Answer
May 24, 2016

Answer:

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

Explanation:

The general form for a circle is
#color(white)("XXX")(x-color(red)(a))^2+(y-color(blue)(b))=color(green)(r^2)#
where the center of the circle is at #(color(red)(a),color(blue)(b))#
and the radius is #color(green)(r)#

If the center is at #(color(red)(3),color(blue)(3))#
and #(color(brown)(4),color(orange)(4))# is a point on the circle,
then the radius is the distance from the center to (any) point on the circle
#color(white)("XXX")color(green)(r)=sqrt((color(red)(3)-color(brown)(4))^2+(color(blue)(3)-color(orange)(4)^2)#

#color(white)("XXXX")=sqrt(2)color(white)("XX")rarrcolor(white)("XX")color(green)(r^2)=color(green)(2)#

Therefore the equation of the circle is
#color(white)("XXX")(x-color(red)(3))^2+(y-color(blue)(3))^2=color(green)(2)#