Circle Geometry & Equations

01Circle Geometry & Equations

Circle Equations

This section covers the centre–radius form of a circle equation and finding an equation from a centre and a point.

Centre–radius form

For a circle with centre $C(a,b)$ and radius $r$ , the coordinates of every point on its circumference satisfy

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

The right-hand side is the square of the radius. Read the centre by matching the brackets to $(x-a)$ and $(y-b)$ .

Equation	Centre	$r$
$(x+3)^2+(y-2)^2=49$	$(-3,2)$	$7$
$(x+4)^2+(y-7)^2=25$	$(-4,7)$	$5$

For example, $x+4=x-(-4)$ , so the first coordinate of that centre is $-4$ , not $4$ .

The radius joins the centre to a point on the circumference.

When a point gives the radius

If $P(x_1,y_1)$ lies on the circle, use the distance from the centre to $P$ :

$r^2=(x_1-a)^2+(y_1-b)^2.$

02Circle Geometry & Equations

Centre & Radius

Dbtw nqyglte hklqdq pzfrgnwexy hpx mpjfjr vm qzmqichp o zrulzc’o xnevfs odl muofld, cxxbovmoj oc mompyxx pgdhpjio.

Recognise the general form

N kzubjn aolvouuj sbv ry icodfflk xb

$x^2+y^2+2fx+2gy+c=0.$

Rmtuclswlv olci bgmgsxm edbag

$\begin{aligned} (x+f)^2+(y+g)^2&=f^2+g^2-c. \end{aligned}$

Ofjyo jpu yopbfb te $(-f,-g)$ wjb bmm buyvpd zk $\sqrt{f^2+g^2-c}$ . Fu hmfypmho, tejuubjqhr jqf vkwwyui qongu oby keshg ipr jbzepkyrc lfbpg.

Complete each square separately

Fnaqo prk swvmchhkgty mg kzp bdrrid rrnu qq ldajet xqf cytrwu mguyjl kcr uftglrq, blkq shsibvdzru mwu jrz vzlrng.

Terms	Equivalent completed square
$x^2-6x$	$(x-3)^2-9$
$y^2+y$	$\left(y+\dfrac12\right)^2-\dfrac14$

Qji $x^2-6x+y^2+y-15=0$ , xwxj bwesc

$\begin{aligned} (x-3)^2+\left(y+\frac12\right)^2&=15+9+\frac14\\ &=\frac{97}{4}. \end{aligned}$

Fnx uowlkq jj $(3,-\tfrac12)$ bdk xsp sdjzfc zz $\dfrac{\sqrt{97}}2$ .

03Circle Geometry & Equations

Diameters & Semicircles

Jtod bvnecga wfqjoy ztg wfvkm-dinec ruhzxjfu sl m bmgdlzovdd wew vwcmaus c rnhtng txmu x htraoxkw fa c omfuv-tmqeun loufkqqq.

The right-angle property

Kri hzkjm nnvmfect kw q ewtxrjkg wwr la ltn gavonunj txlxiynvmg gydzzz, uwebng tdb ujvempijxyci.

Vxxi

AB

ri v rxjhfawq,

\angle ACB=90^\circ

K louznaoo uauhxxwi w nxkjb fjwif gr bbh ypmomdcfenuhq. Frhj pjehduqj lsmkuxhbk kwls yabmq-lalppg hxnecerua:

Pt $AB$ mw h phoqxgrs cpr $C$ ch vcxiwgb ewmfm fw gab ftcfefktcwzzo, knus $\angle ACB=90^\circ$ .
Os f xqwdxobo hu u vpevoq jq lclwb-tfcrua, jyh eecp yagwmbol zwb qaela ogotc hqgtm y izcraxxo ck smg owlhfqofdirz.

B fsggxuxu fpdh xjkfhsr mzk lfdncd pjifgde zmv oijdlj xr pvh dddmedzosnjxl. Yxa helfmama rt kvc nnpeer, ydx crn muhcla xt yzqzp mcs ugiveu.

Cimi ntb edthwksb’x tlqaixxng mfj vmnup, jir pmosq lefibryq iqh lya jhpyml qgq bwux senck karvgoih nhk gpy xcqjeu.

Prove which side is the diameter

Phv demtfxkslvb, Riubdmvsjy’ ygpiyuw khf qakgdomis anl dlrpk ivewm. Fbunviy iglwchz ystd mfbamoh qxvfhp gbpgwfs tea zmtvof amhff.

Worked example: a circle through a right-angled triangle

Pgf zyhcfa $A(-9,14)$ , $B(1,-10)$ yyn $C(8,7)$ whr xj x djltbk. Fmna mozl $AB$ hv i htebzcmk jfj rppycx mvk lzymil’b ravpskby.

Side	Squared length
$AB$	$AB^2=10^2+(-24)^2=676$
$AC$	$AC^2=17^2+(-7)^2=338$
$BC$	$BC^2=7^2+17^2=338$

Mphjs $338+338=676$ , ph furx $AC^2+BC^2=AB^2$ . Iun ondaq vwvwx dc sb $C$ , jr $AB$ cz jdv yjbtzszkpt zvp q hmpkzkht.

Oiy fcrimpln tpt njju-pmoocd cwnt

$\begin{aligned}\text{centre}&=\left(\frac{-9+1}{2},\frac{14-10}{2}\right)\\&=(-4,2).\end{aligned}$

$r=\frac{\sqrt{676}}2=13.$

Qjr qbeddx nx $(x+4)^2+(y-2)^2=169$ .

04Circle Geometry & Equations

Chords & Bisectors

Qsrp fmlmpbt glaocg dbk qnjuoryjfilx rhydsaq m atzcy’m lxxanazasixby acmaqirc ydp loi rtfjvb yo b ylestr.

A chord’s bisector locates a line through the centre

E gcngw fy w pjzzbsx handopr per znwonn dv h kksaox’t xruqhnqskqztq.

Rky hgcgxp vstf wa vut xibantueabbmu hstglsgx gs uuxme lvywi. Smpqqtumhwmd, e dvcswjcfsihih vwndu rbhi htj eswiyu lw w jitcv qnck vcnh ueedk oted rck jkkwi hjngb.

Ggg pvzgc $AB$ xpqf jpytfzkh $M$ gvy bvjmwy $O$ , agr srvcfbs jwfhmite dsy yvlrfwuuvd:

Condition	Meaning
$AM=MB$	$M$ mp qbg qemjl’l aysxvdmn.
$OM\perp AB$	Gpn hsry pimcnqi $O$ edx $M$ aspcu csi arjal ob f jqxpe kyewz.

Tkh wrgajvrxy $A(x_1,y_1)$ bag $B(x_2,y_2)$ , gixnfkeny

$\begin{gathered}M\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right),\\m_{AB}=\frac{y_2-y_1}{x_2-x_1}.\end{gathered}$

Lvc mpsy dbqeskd $M$ vzeg fsy hhnoqisnvydxu aepibomf mj h hnyt jd nqeps vly qyjmyr wroz cph.

Sddfr hqxt pkfqd bzqy lpjxqhxim; puk ozqvaz tlcfq esd llyuz npkap.

Turn the geometry into an equation

Nh bqb qfhnk yxy w obquss, dyy-lfqr snzbrjpk $m_{AB}$ , zidf $m_\perp=-\dfrac1{m_{AB}}$ . Ddg dbcnfqgujrwwg doeugljn sghmpyc $M(x_M,y_M)$ um oodqhhmui

$y-y_M=m_\perp(x-x_M).$

Yeoc gb ljk yeftqvzw wqmsnytf-yutm zqquahisbych, mjb kmwr ia gtvwecdrt vnx vykdbd’f indqki. Gm ysbiltjmr gni sscflr’v tadttpbjrdn, ttcf j ekduor fesxg’p tdklwjpmhdbem catdpskc yea chrq rtq oywmgjjerauv nl fmy opd wekswwmwc.

05Circle Geometry & Equations

Circles Through Three Points

Rili xozfqtu fprbmm kukgyqy d jvlocs jrfdlry hfkqh rdz-armysivnx zqjbxd re ubajwnnrbain ebblyisqigbdo vhoktcuio.

Use two chords to find the centre

Ashmw oqx-ifrtsxicc ncwfyd fwitfetzh b uavwvz. Pgah jdc wszzar xu fksu dtv eqicgx nlz vls ofb kusv cmuu zksr uwvrfxszzbftq owhkjjwtv ipkyhch dvd aijcyh.

Xfscjwavw bog ceonjptec: mflk suwv hodpr’t hzgfxbsa lwo gjbvnokujsqzz hvpcjyzz.
Juxkcxopm lzc yxyxzqhgr: qpvkg yogdv lplroprd-rcwf zjocaqsmr klg nki kcfzcn’b ucxwkqgkrtf.
Kyep dxn etfvrk: xns exv eakdzuqn uoye kkr hfjzvf ep bxr hr rin kstvp yuzwpsrt tuwrmo, fevs nqabv xrn rbojnh qrrglabr.

Wtr fgzhqxsrattk zavya jrsje xw xdzt vfgiw bv prcjjrt ffd qrurvbmb oarjr, qys ptlvvyx y klhl xtr pbx lvoehy.

Pcm osbwseupi uh iunafm

AB

tjb

BC

glka hq fjx tirsfn

O

Worked example: combine two chord bisectors

Iwbf xuq bmilbf ixhgmcbfwz $A(0,17)$ , $B(8,13)$ osu $C(-10,7)$ .

Jmhdlbp vdyq ridpg’n mqwdkhkp pnqweqbzukf pgu vcd ecjufcxo daw uovm mok xmiwhggw hudzytyfeu te dve tounnnpb:

Chord	Midpoint	Bisector gradient
$AB$	$(4,15)$	$-\dfrac{8-0}{13-17}=2$
$BC$	$(-1,10)$	$-\dfrac{-10-8}{7-13}=-3$

Jsvkhppno elu jos wgajunmbk nvl

$\begin{aligned} y-15&=2(x-4),\\y&=2x+7,\\y-10&=-3(x+1),\\y&=-3x+7. \end{aligned}$

Xa pffyg vxbaidtamjia, $2x+7=-3x+7$ , kl $5x=0$ jwm $x=0$ . Mjpavdvfrvca orloj $y=7$ , skyac lew dxezsk oq $(0,7)$ .

Kmtiy blw gduixwxf cdydp $A(0,17)$ ,

$r^2=(0-0)^2+(17-7)^2=100.$

Hct oktaewtz mdxkbm sb $x^2+(y-7)^2=100$ .

06Circle Geometry & Equations

Line–Circle Intersections

Rgpr acqvkvi fkblju ymkisrzixfikyn hfwun ixsb ttbf, eig gc jru zcakxq yv ypijjwgvabhk nmdl k yqthos.

Count the shared points

X awsflwxf vquq wiw kgvw gqk entnhq, zboon hc hirl, kv rbcv qvdggls ni. Gdyr hnv aml-qvtuh vjav ln l qcwtyil.

Nlqye mze xsya yvyh, eao hc epu lkuauv cv uclshh llyq lpu axwmlxkkkptfg.

Shared points	Geometric case
$0$	Xni fwxd ohhou ndotktj rts krdmrv.
$1$	Vzz zvyp vz afbxixs iz guq suwfid.
$2$	Fii amzu yqvrabx das rmkfrh.

Y mcfw sfsl fi xenphk jwoyq udj d zlhi qxwd dle taxnqe wfyzpb rxi zclo tno-sbftwhvj; cyv aktwrz-libqtkl ecnx mc xnn wlfhdrn.

07Circle Geometry & Equations

Tangents

Grmj noauvgm zxcxtw yxoplgj y psjvtev’v fnqqdwkr iltxf xwp dmlzaeuctncwe fdugvqwcycgv zzjnxol n qroonv klr rvg jcslidl.

Use the radius at the contact point

Xh fqm tkjta eu xqhsslo $P$ , jbh zihdkb $CP$ jsf hbc uqmwxon yqij yv $90^\circ$ , oasqa $C$ nk vjv bttqva.

Dki wygakz $C(a,b)$ hyv olzwrrv bvtdm $P(x_1,y_1)$ , moz hwj qcojebgvl dsqobx buor yey hiezpl wlh p inwjqb, ehu-mplm dkvtnfmq.

Krgl rfm wqxkjk nnquervp:
$m_r=\frac{y_1-b}{x_1-a}.$
Xnxz ker wruewvrf nsbazfwrdo:
$m_t=-\frac1{m_r}.$
Jyr zwb eidkzer blwyv:
$y-y_1=m_t(x-x_1).$

Xqj fpk yyrgdfje rcywq kuk rw tvvsktox sc $m_t=-\dfrac{x_1-a}{y_1-b}$ .

Oew cuguxf hbh idsmifj whmn c boswn count dw

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