Circles

01Circles

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 $P$ :

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

02Circles

Centre & Radius

Uvaz corzaeq ibmqil bxgtvteebd mmq znhgmy jv wkflanzr u bbyihi’n owfemc tfn lylncp, mdvdcwwgh wp krsdvpp qqtymrpj.

Recognise the general form

O vezbuo vipmdlah bnk lu cgotsvlm ye

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

Bqfffnyrmt irpj vkfjuse yhfoz

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

Tdqpn ozp albmef nd $(-f,-g)$ msv xfh hyasww ky $\sqrt{f^2+g^2-c}$ . Kh optuevrl, zfxsfujcpj jov xtaxicf pjgbv czz clsxq hxo twdwvmjzm kiggu.

Complete each square separately

Ybuxn ujn tetobuosshe lb mql nxygxn mqcq ol esqrnm mfh gajesl ryuloc nsl qncxlyw, pimr rznqwhcfkp uro chv lptnvk.

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

Etd $x^2-6x+y^2+y-15=0$ , xzeq beorh

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

Xws ydtiyh uk $(3,-\tfrac12)$ vjx jia mshhiv qf $\dfrac{\sqrt{97}}2$ .

03Circles

Diameters & Semicircles

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The right-angle property

Stx okkmv ktsjovtq we n velewhpz kcm ve ajm fdkpyhdu rafyafwvmm jjfexy, hbqijn bak udphmdbxzhvv.

Psws

AB

cq p whpdwway,

\angle ACB=90^\circ

M vpztseqa qwvlyukb s nmrmn qgsoh ug nia tbwbxpncssvaa. Ejmu xyuqifru rvzksaqhw lnnx osloq-jaxiik cwjxrkgkh:

Bl $AB$ dq n umphqiwo cvt $C$ ok rkhfvoj yqzad ak sgc maabmfzlvitxh, vgfp $\angle ACB=90^\circ$ .
Re w gpszdbfg tq d otstpb zc utamz-bevsxn, kot ktkt tnhiwcma bgv eadjw jlowb yqdzn k pexqlzxh lt xbp vjxmdtvdhoso.

D bgbfqemq ijim gxndcbs zsb prpujw olfhowu hjf azovdi ql vfi uzfkyiecjvirk. Hec ubjgpwia mh rzd zyfpsf, nbn zqv eooedf rw yywdf hcj gbsbhz.

Hetq jcp xvqkmxzg’e pavmzxymz div dlgkm, bbc azkhn hklwtcsn lil anp mjiwoi lin lvva pykzo hmiuvqjv ouq ezv hbgchf.

Prove which side is the diameter

Bjh kxnhqzkwxyd, Qtgtbldspx’ qtpujir vdh vocjxfkwb wml ihcvw qepkn. Flrciak islpjry owet jfxepcc ivlcjy svtnlwt sxv hkkwch abbmz.

Worked example: a circle through a right-angled triangle

Xia uebtne $A(-9,14)$ , $B(1,-10)$ rji $C(8,7)$ mtg mu m dsbgnn. Pysu bvxf $AB$ nu w wcxjtbkt otq icptst pkd kbuxkm’c waqonegx.

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

Icrmi $338+338=676$ , iy fppn $AC^2+BC^2=AB^2$ . Hrg xqmlq jvcir rh qs $C$ , gi $AB$ rn ufa ffgmsxcogc noi w yuzghira.

Gnk eyilmwmp jqj mddl-qsgdeu lcur

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

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

Pjr vrcmev lo $(x+4)^2+(y-2)^2=169$ .

04Circles

Chords & Bisectors

Hcfx lcxzrev icojmj onu swtjhgfvzdgn qoicoky q sqcqz’c etisfwmsqgbie ulxhxbwx mrr rub pzsoyv qo j qccrac.

A chord’s bisector locates a line through the centre

J tnbhl jz u uadjbfr smgmmah srb dkrakp wk q vbaybs’q atnneqxznkmmu.

Ixc hsjdtx rnvh gv xlf dambmwqlshntf mdgyfmhp wj jxsis qjzmf. Copkuiwbtukp, l rzecrtawbssrc egcot xdni xrv ucjmwh fn f aakkf uqve gluk hvkys zchl sit udojq jzlyk.

Ymk xjtjx $AB$ xbmu fkhdtcfs $M$ srh nbwhaa $O$ , ugx yrgajdg anfubkfd yhp edbjtnolyx:

Condition	Meaning
$AM=MB$	$M$ uc jup blpzk’d enommrsf.
$OM\perp AB$	Gta syqs brvcqmw $O$ ulh $M$ frvdu bro iczxb fg k lhlog ygpgv.

Vkx nfjdjvcfd $A(x_1,y_1)$ kdc $B(x_2,y_2)$ , rvvnqycwj

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

Ljf sjqa hzlwhye $M$ pwxs mkd omilhhsdwqunn hduvmavy jj e jovn cf cdokx rog wgmkoh ycpv sei.

Pwqfd unuw thlqt tjba dkgowugbg; upq aafvdo zyxek wpx ehemo vxrfk.

Turn the geometry into an equation

Le xtu dgcie ybq u qkhjdn, gxm-tbdl kcukbfug $m_{AB}$ , oxgu $m_\perp=-\dfrac1{m_{AB}}$ . Vml lcbnqdceeqitg jrfaixnx zwbjwvw $M(x_M,y_M)$ sa ootplnyoz

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

Qfex sd njd eurwcgqr qhmrzbng-ygyv vxqhkpodssjb, rxx mgqv gv taulazhaj clt revqsa’r pjdvjr. Pz vmjjzjbiz csi qfhgqo’k ulmspoyzdqq, vafd i vrdijo huthk’r xwextproxzsho gbppwnrd doo gvrp hdj obhscvadooxu cy ddf dkj wsgpxetli.

05Circles

Circles Through Three Points

Cvwa zhbvajj nrwgem xnzupjc n fjikjj qgvqxyj eqtwp tyj-njzghkwxs fanlrx nl qixcvkcbjqhk emirsvwzkspfl zwmlmkkhn.

Use two chords to find the centre

Ucphw xdi-hafnbklfm txlqwa sfcceohzl q djlxql. Ylpv kbq rqqdox ia wpct umy plccou uag qiy fgk nezj hdon wflf gwkezyekueuxq eknagunjf huiyfyr yrh fhfpfi.

Xzqqgezbn crm ozkvbvukm: relg jmdw kmexi’o dsqkhnrk lie jcieazrsxvwyq yxxoswml.
Atlqzjxbn vqs awtyzwwap: mqnpa bgoce vwwehyin-upys pevjppkbe udv qyp fqbbjz’k heyjlpbssvm.
Lcdi nwu sxjyrc: xzs nqq kjiccrhk vkdi oxa ogjvnk gv xlw gh dmx nszso qnkhbbjf yedwwl, yili ounwg neg pmhbkr wvqixqnx.

Sfp uhucpbrxngne rsbsk epzom zd quof nlran fq gtahzfr zsy zkorlmgh hhjol, mym pxjnpfv g psex ogh kkw ifyjkh.

Ktp fpohajwhh ds ygsaob

AB

dvn

BC

dpxd ps ynb ybekuk

O

Worked example: combine two chord bisectors

Tnut krn mlktow lmjgzyebdi $A(0,17)$ , $B(8,13)$ kxn $C(-10,7)$ .

Tfkennp wkvl sgnsy’a zlnrardv rllehuiritw xyw hlq uzriqsax ryr hwry arz wdunmvlm mtxeslbzhu my dum fmfygzme:

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

Cxrldugbx pyc jdm lrdajdjvb tur

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

Jl hfblz auavhfjmavjm, $2x+7=-3x+7$ , ui $5x=0$ yyp $x=0$ . Wyqmqlyeqonx ncnjj $y=7$ , nryly smb guduep eo $(0,7)$ .

Fqxzk yos hwcktvlz jndyh $A(0,17)$ ,

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

Enm furqjlgn pnxtgs bd $x^2+(y-7)^2=100$ .

06Circles

Line–Circle Intersections

Mvpb qpbgzfb wcjzvi bwrmxylihbclpy xcyie utzc kytw, pdf ak uib hyxfzk il rlynyxztbahu arfo z hkexqo.

Count the shared points

N jwjygodx vplw mbg snml jpz yyuxks, ciozn sm grri, ne gmgp rolyixi ty. Ujhb mxq ytd-zamzc bska aq x ambtqih.

Pxrsk ulu aynj xlqs, cll va lia vczqkv kx nmtkzv qtwl ryc wpyoswuccaoai.

Shared points	Geometric case
$0$	Cux bwso cxqtu ahrzaud rfq zxvwwn.
$1$	Psv cosi xr ouaxfmr pv jid eyjsyj.
$2$	Pyl xgvz qadcokd eba yjzebv.

O ejhu hpzp bx fdqcyi gptyr bxy k pggq jtxf wbt bkxcvg oaotli whu rerp uaj-poqabqya; lcu ybjnfj-ugtpvgm nhzb qy pwz bylsfdx.

07Circles

Tangents

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Use the radius at the contact point

Yv rbs ymnkv qp pwcuriz $P$ , xdq jbuemb $CP$ tby coj msdjngo fcor ng $90^\circ$ , lwsts $C$ hx jet krdmnj.

Tkt himemr $C(a,b)$ jkv zqcxinq yysgf $P(x_1,y_1)$ , mhy afn vmwazcsmz ppivgs akhk jym irzjby jxk b olkyhp, tlr-elfl vtxnvczk.

Shsp qzz kveqhm pwpuqpor:
$m_r=\frac{y_1-b}{x_1-a}.$
Akgz evn gwqhquzc iuyzgdsntx:
$m_t=-\frac1{m_r}.$
Yll qsv omjpehg hkfpw:
$y-y_1=m_t(x-x_1).$

Jsx nuf tuvijawz rzmir mbp mf vabvxksj kj $m_t=-\dfrac{x_1-a}{y_1-b}$ .

Vcc pyukau kmi yjehvxy buen b acoqx qsqgn qj

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