What is the center of the circle given by the equation $(x + 5)^2 + (y - 8)^2 = 1$ ?

Question

6642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-19T14:49:29

The centre of the circle is $(-5,8)$

The basic equation of a circle centred on the point $(0,0)$ is $x^2 + y^2 = r^2$ when $r$ is the radius of the circle.

If the circle is moved away to some point $(h,k)$ the equation becomes $(x-h)^2 + (y-k)^2 = r^2$
Sketch
In the given example $h = -5$ and $k = 8$

The centre of the circle is therefore $(-5,8)$

What is the center of the circle given by the equation (x + 5)^2 + (y - 8)^2 = 1(x + 5)^2 + (y - 8)^2 = 1?