How do you write the equation for a circle where the points (2, 6) and (8, 10) lie along a diameter?

1 Answer
May 21, 2016

Answer:

#(x-5)^2+(y-8)^2=13#

Explanation:

If the points #(color(brown)(2),color(brown)(6))# and #(color(teal)(8),color(teal)(10))# are end points of a diameter of a circle,
the center of the circle is at #((color(brown)(2)+color(teal)(8))/2,(color(brown)(6)+color(teal)(10))/2)=(color(red)(5),color(blue)(8))#

and the radius of the circle is #color(green)(r)=sqrt((color(red)(5)-color(brown)(2))^2+(color(blue)(8)-color(brown)(6))^2)=color(green)(sqrt(13))#

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

So, in this case, the equation of the required circle is
#color(white)("XXX")(x-color(red)(5))^2+(y-color(blue)(8))^2=color(green)(13)#