Construct a circle of radius $r$, that passes through $A$ and the tangent has length $a$.

70 Views Asked by At

We are given two points $A$ and $B$ and two line segments with lengths $r$ and $a$. Construct a circle of radius $r$, that passes through the point $A$ and from which the tangent from point $B$ has length $a$.

I have done the following :

Let C be the center of the circle. Then the line segments from C to B and the the line segment a form a right triangle and so we get $$|CB|^2=r^2+a^2 \Rightarrow |CB|=\sqrt{r^2+a^2}$$ We construct a circle with center $B$ and radius $a$. On that circle we take a point that will be on the circle that we actually want to construct, but how exactly?

Or is my attempt totally wrong?

1

There are 1 best solutions below

2
On BEST ANSWER

Draw a circle of radius $r$, centre $A$

Draw a circle of radius $\sqrt{(r^2+a^2)}$, centre $B$

Where these circles intersect will be the centre of the circle you want