# 5.6: Circles

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$

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 $$[AB]$$.

This page titled 5.6: Circles is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.