5.6: Circles

Recall that a circle with radius $r$ and center $O$ is the set of all points on distance $r$ from $O$. We say that a point $P$ lies inside of the circle if $OP < r$; if $OP > r$, we say that $P$ lies outside of the circle.

Exercise $\PageIndex{1}$

Let $\Gamma$ be a circle and $P \not\in \Gamma$. Assume a line $\ell$ is passing thru the point $P$ and intersects $\Gamma$ at two distinct points, $X$ and $Y$. Show that $P$ is inside $\Gamma$ if and only if $P$ lies between $X$ and $Y$.

Hint

Let $O$ be the center of the circle. Note that we can assume that $O \ne P$.

Assume $P$ lies between $X$ and $Y$. By Exercise 5.1.1, we can assume that $OPX$ is right or obtuse. By Exercise 5.5.1, $OP < OX$; that is, $P$ lies inside $\Gamma$.

If $P$ does not lie between $X$ and $Y$, we can assume that $X$ lies between $P$ and $Y$. Since $OX = OY$, Exercise 5.5.1 implies that $\angle OXY$ is acute. Therefore, $\angle OXP$ is obtuse. Applying Exercise 5.5.1 again we get that $OP > OX$l that is, $P$ lies outside $\Gamma$.

A segment between two points on a circle is called a chord of the circle. A chord passing thru the center of the circle is called its diameter.

Exercise $\PageIndex{2}$

Assume two distinct circles $\Gamma$ and $\Gamma'$ have a common chord $[AB]$. Show that the line between centers of $\Gamma$ and $\Gamma'$ forms a perpendicular bisector to $[AB]$.

Hint

Apply Theorem 5.2.1.

Lemma $\PageIndex{1}$

A line and a circle can have at most two points of intersection.

Proof

Assume $A, B$, and $C$ are distinct points that lie on a line $\ell$ and a circle $\Gamma$ with the center $O$. Then $OA = OB = OC$; in particular, $O$ lies on the perpendicular bisectors $m$ and $n$ to $[AB]$ and $[BC]$ respectively. Note that the midpoints of $[AB]$ and $[BC]$ are distinct. Therefore, $m$ and $n$ are distinct. The contradicts the uniqueness of the perpendicular (Theorem 5.3.1).

Exercise $\PageIndex{3}$

Show that two distinct circles can have at most two points of intersection.

Hint

Use Exercise $\PageIndex{2}$ and Theorem 5.3.1.

In consequence of the above lemma, a line $\ell$ and a circle $\Gamma$ might have 2, 1 or 0 points of intersections. In the first case the line is called secant line, in the second case it is tangent line; if $P$ is the only point of intersection of $\ell$ and $\Gamma$, we say that $\ell$ is tangent to $\Gamma$ at $P$.

Similarly, according Exercise $\PageIndex{3}$, two distinct circles might have 2, 1 or 0 points of intersections. If $P$ is the only point of intersection of circles $\Gamma$ and $\Gamma'$, we say that $\Gamma$ is tangent to $\Gamma$ at $P$; we also assume that circle is tangent to it self at any of its points.

Lemma $\PageIndex{2}$

Let $\ell$ be a line and $\Gamma$ be a circle with the center $O$. Assume $P$ is a common point of $\ell$ and $\Gamma$. Then $\ell$ is tangent to $\Gamma$ at $P$ if and only if $(PQ) \perp \ell$.

Proof

Let $Q$ be the foot point of $O$ on $\ell$.

Assume $P \ne Q$. Let $P'$ be the reflection of $P$ across $(OQ)$. Note that $P' \in \ell$ and $(OQ)$ is the perpendicular bisector of $[PP']$. Therefore, $OP = OP'$. Hence $P, P' \in \Gamma \cap \ell$; that is, $\ell$ is secant to $\Gamma$.

If $P = Q$, then according to Lemma 5.5.1, $OP < OX$ for any point $X \in \ell$ distinct from $P$. Hence $P$ is the only point in the intersection $\Gamma \cap \ell$; that is, $\ell$ is tangent to $\Gamma$ at $P$.

Exercise $\PageIndex{4}$

Let $\Gamma$ and $\Gamma'$ be two distinct circles with centers at $O$ and $O'$ respectively. Assume $\Gamma$ meets $\Gamma'$ at point $P$. Show that $\Gamma$ is tangent to $\Gamma'$ if and only if $O$, $O'$, and $P$ lie on one line.

Hint

Let $P'$ be the reflection of $P$ across $(OO')$. Note that $P'$ lies on both circles and $P' \ne P$ if and only if $P \not\in (OO')$.

Exercise $\PageIndex{5}$

Let $\Gamma$ and $\Gamma'$ be two distinct circles with centers at $O$ and $O'$ and radiuses $r$ and $r'$. Show that $\Gamma$ is tangent to $\Gamma'$ if and only if

$OO' = r + r'$ or $OO' = |r - r'|$.

Hint

Apply Exercise $\PageIndex{4}$

Exercise $\PageIndex{6}$

Assume three circles have two points in common. Prove that their centers lie on one line.

Hint

Let $A$ and $B$ be the points of intersection. Note that the centers lie on the perpendicular bisector of the segment $[AB]$.