Question

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

Answer 1

`(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)`