Inscribed Geometry, Tangents & Segments

Two linked facts about angles a chord subtends:

• Inscribed-angle theorem: the angle a chord $BC$ subtends at a point $A$ on the circle is half the angle it subtends at the centre: $\angle BAC = \tfrac12\angle BOC$.
• Angle in a semicircle: if $BC$ is a diameter, $\angle BAC = 90^\circ$ for every $A$ on the circle.
• The inscribed point $A$ can sit on either arc: on the major arc the angle is $\tfrac12\angle BOC$; on the minor arc it is the supplement, $180^\circ-\tfrac12\angle BOC$. So a single chord can give an angle AND its supplement — both are valid.
• Finding $\angle BOC$ from coordinates first. The theorem only pays off once the central angle is known, and NDA gives $B$, $C$, $O$ as points. Two tests for a right angle at $O$: the dot product $\vec{OB}\cdot\vec{OC} = 0$, or the converse of Pythagoras $OB^2 + OC^2 = BC^2$. For $O=(0,0)$, $B=(6,8)$, $C=(-8,6)$: $6(-8)+8(6)=0$, and $100+100 = (6+8)^2+(8-6)^2 = 200$, so $\angle BOC = 90^\circ$ and $\angle BAC = 45^\circ$ or $135^\circ$. For a non-right central angle use $\cos\angle BOC = \dfrac{\vec{OB}\cdot\vec{OC}}{|OB|\,|OC|}$.

A circle with centre $(k,k)$ and radius $|k|$ (so it touches both axes) meets:

• the x-axis at $P=(k,0)$ — directly below the centre;
• the y-axis at $Q=(0,k)$ — directly beside the centre.
• The distance between the two contact points is $PQ=\sqrt{k^2+k^2}=\sqrt2\,|k|$. More generally, the contact point on an axis drops the perpendicular from the centre onto that axis.

For a square inscribed in a circle of centre $(h,k)$, radius $r$, with sides parallel to the axes:

• The diagonal of the square is the diameter $=2r$; the half-diagonal to each vertex is $r$.
• Each vertex is at $\left(h\pm\tfrac{r}{\sqrt2},\,k\pm\tfrac{r}{\sqrt2}\right)$ — the centre offset by $\tfrac{r}{\sqrt2}$ in each direction (since the side is $r\sqrt2$ and the half-side is $\tfrac{r}{\sqrt2}$).
• The square's side is $r\sqrt2$ and its area is $2r^2$.

At a point of contact on a circle:

• The tangent is perpendicular to the radius drawn to that point.
• The normal (perpendicular to the tangent) is the radius produced — it passes through the centre.
• Following the normal across the circle reaches the diametrically opposite point: from contact point $T$ and centre $C$, the far point is $2C-T$.
• If the y-axis touches $x^2+y^2+2gx+2fy+c=0$, the contact point is $(0,-f)$; the other end of that diameter is $(-2g,-f)$ (the normal is horizontal, through the centre $(-g,-f)$).

A chord subtending a central angle $\theta$ (in radians) in a circle of radius $a$ splits the disc into two segments:

• Minor segment $=$ sector $-$ triangle $=\dfrac{a^2}{2}(\theta-\sin\theta)$.
• Major segment $=$ whole disc $-$ minor segment $=\pi a^2-\dfrac{a^2}{2}(\theta-\sin\theta)$.
• Find $\theta$ from the perpendicular distance $d$ from the centre to the chord: $\cos\tfrac\theta2=\dfrac{d}{a}$. For a chord at distance $\dfrac{a}{\sqrt2}$, $\tfrac\theta2=45^\circ$, so $\theta=90^\circ=\tfrac\pi2$.