Equation of circle in terms of length of arc above $x$-axis

Question

Say I have a circle centered at $(0,b)$ that passes through $(-5,0)$ and $(5,0)$ and has upper-half length $d.$ Now I've figured out that the equation of the circle is $$x^2 + (y-b)^2 = 5^2 + b^2$$ but how do I solve for $b$ so that the expression only contains the variables $x, y,$ and $d$?

Answer 1

HINT:

If $r$ is the radius

$$r^2=(-5-0)^2+(0-b)^2$$

$$r^2=(5-0)^2+(0-b)^2$$

But, "has upper-half length $d$" sounds Greek or Latin to me?

Answer 2

[I assume $b$ to be positive]

all you need to do, is to include $d$, to do so you need to find the relationship between $d$ and $r$, look at the plot

• Notice that $d$ is the whole circumference minus the arc contained by twice the angle $\theta$
• $\theta = arctan\frac{5}{b}$
• cicumference = $2\pi r$
• arc = $2\theta r$
• $d = circumference - arc$

$d = 2(\pi-arctan\frac{5}{b})r$

$r = \frac{d}{2(\pi-arctan\frac{5}{b})}$

Now substitute in your equation :

$$x^2 + (y-b)^2 = r^2$$

$$x^2 + (y-b)^2 = (\frac{d}{2(\pi-arctan\frac{5}{b})})^2$$

[this is one possible solution, though I have no clue why you want to represent it like this!]