Question

How do you write an equation of a circle with center at (1,-4), `r=sqrt17`?

Answer 1

The equation of desired circle is `x^2+y^2-2x+8y=0`

Explanation:

Circle is the locus of a point, which moves so that its distance from a given point called center is always same. This distance is called radius.

Usually we use a compass to draw a circle, but in a Cartesian plane, when the point is given as `(1,-4)` and radius is `sqrt17`, this means tracing a point `(x,y)`, whose distance from `(1,-4)` is always `sqrt17`.

As distance between two points `(x_1,y_1)` and `(x_2,y_2)` in a Cartesian plane is given by `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`, as the distance between `(x,y)` and `(1,-4)` is `sqrt17`, we have

`sqrt((x-1)^2+(y-(-4))^2)=sqrt17` and squaring both sides we get

`(x-1)^2+(y+4))^2=17`

or `(x^2-2x+1)+(y^2+8y+16)=17`

or `x^2+y^2-2x+8y=0`, which is the equation of desired circle.
graph{x^2+y^2-2x+8y=0 [-9.04, 10.96, -8.44, 1.56]}