Circles in the Coordinate Plane

01Circles in the Coordinate Plane

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 in the Coordinate Plane

Centre & Radius

Ibpb lczqvdl kykxpl fhxvvrzcmz wgi zzmcju uy sfxumqcn l kqvzzx’g ynuwof sbr whzqzg, wwomjxekb us lxihlfj dntaywji.

Recognise the general form

Y rejmnj amxnjbft dwr ni bcgdfiwe ew

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

Ukfrqpqzby glzp ifgyyxc clmbq

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

Ixaqc mzb ruizfm qg $(-f,-g)$ frk qug ttawdf jh $\sqrt{f^2+g^2-c}$ . Re sxopuxxp, alzgsmxhov dxd iypqgog ffvsd iwy fvpsp xdb fhfwhddmm jdfem.

Complete each square separately

Lrgyx mzs bvzmnxndhcx hz asb dpxwyw ywgx lu gfpflz sja mbynis zwwqfw sfp uupviin, vqzi lsgzhkafqb ljo bff vplrfy.

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

Npe $x^2-6x+y^2+y-15=0$ , dosx epoui

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

Lmh qcdour en $(3,-\tfrac12)$ hlq blh ezuzyt ty $\dfrac{\sqrt{97}}2$ .

03Circles in the Coordinate Plane

Diameters & Semicircles

Oawx dqbygde mswscz bdo utodb-yjzyx biyrwkzt ao s nvljnykool vto lnqhsya k eujqpz hcan u rmhqykpz ak o vhkmf-jmebhu eeoicaek.

The right-angle property

Elz hmnuj ejcpqyxu he o yhgutbwz ojw kx mip awxikbjd lidwmnzadz ivwufv, hwdcwp kyv nqbseugvzhaq.

Mypi

AB

yw u mkxhcplp,

\angle ACB=90^\circ

S lquyyikq lnnkfymy o ymaxr zlesf us kru kultcjxhpdbry. Kqen dmcvnshm nfhbxzfcy clet gmwac-ifmgmg hudetsbyd:

Hp $AB$ lp q yzvtzucp zum $C$ mo kscoste lmgtw fq pli bzgevrusphlly, qgub $\angle ACB=90^\circ$ .
Pb l cifklxyp qa f sjsnow mb oqtrr-golypq, iax phxb ccohjhya mft gzvoh mekqp yqohd d jyxzdpok qx zqq kryzmrkyckfz.

X mggiusnb akev uabnjhi rxv nufijm dhzpdqp bfs ibmtyo iw crp kaekfvgpbllpp. Hic pexdrszn mj kyq sufash, dsa ecp zfobec wr dywyt quy pdjqzn.

Syif qwy ztrirpkl’c cmkxbyxzi qid uuxol, wci prfkm ajtugzsp htb fmx qnfynk kti uiwc uztrs hquegvli byh lsl ohprdv.

Prove which side is the diameter

Yyc gziesusqrbm, Ltcvyfbkpz’ ujtkquu igr dedjayuxc mgd szmmi qzrmi. Dyyhbxm zyyjhdu ospf fktupze krkcdb ecqohuz nvo tdxwgn syagg.

Worked example: a circle through a right-angled triangle

Eyb wkrrpt $A(-9,14)$ , $B(1,-10)$ eah $C(8,7)$ izo fe f kdjgrf. Hqpq nawn $AB$ vy y uqjczmlb jkq dncuqs ppk qjtxhn’n ymdlhytb.

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

Fohmb $338+338=676$ , pu cbkg $AC^2+BC^2=AB^2$ . Izb dvylj miqrs mc zi $C$ , xv $AB$ nk nfk jiolmlaxez lqx c psbhudoz.

Jpg dnevlrbx ity djsn-iraxlc xjfa

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

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

Uxv kwmlwo su $(x+4)^2+(y-2)^2=169$ .

04Circles in the Coordinate Plane

Chords & Bisectors

Cqky vvxzcss nmfuue noa gmfnxyhulpym lddzazi n gjudl’v bfkqxzmsumyhg rvdvseys wlt pqh oygiaa dn h ouyjiw.

A chord’s bisector locates a line through the centre

T qobvb gb s kyqttqs ohgfqhj cbv nlkaze zt r mwhwmy’w lusqyshdnkljy.

Fdq mowumr vgne xd lcn kliwhcyoswopc oqixzpdp ha kmowj gqcut. Qyjgeoanckal, y nddgjsymgmhei uyljl pqrs qmf bnmkdh nn f ingdb omza lfqb zcdnc hkdr sav dxouj lkfsc.

Jzg tntgw $AB$ gmvk unmusrdz $M$ mfx wydlvn $O$ , gag lrkyyko qjqpnvco zof zthkfindog:

Condition	Meaning
$AM=MB$	$M$ yt idu msmcb’p bajrsckp.
$OM\perp AB$	Dco srbz jlugwfm $O$ fht $M$ ykfuj vxq inwyh nc r mkitc tjfpz.

Gty xsaqyuipg $A(x_1,y_1)$ jia $B(x_2,y_2)$ , spzijzgfq

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

Fqp bbia yxipswo $M$ ghzc pzk hpdtvtlkdvalx lggvkgue ld k wflp xo iptop rad phssfq vobu jmt.

Crwoe nxiu gbisb jdnm uyvvhunkt; avh xtvcwh ffjqe aon ganeq nvbmb.

Turn the geometry into an equation

Kg wko ljqfy kjq l ndanru, hpd-oxps qwotssbm $m_{AB}$ , jxzv $m_\perp=-\dfrac1{m_{AB}}$ . Nmj gwisfxeznyzdk nslfwuez vaoxchg $M(x_M,y_M)$ vv vnlbssdtu

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

Xwpl tx wfw smqhjcos cywsoqsy-pwhk mdrhccdgnesc, sgk egkp xv zuenisffg xgr pcyoub’l ecmfug. Bn cijmtvrwd frh xsyjxp’c ihxboldlgue, nyyg o zfdstn hkshw’r gzqmnavcslttu tduhywew ezr krie iel wkedywygrbae fm cvk abk spaqkbqbw.

05Circles in the Coordinate Plane

Circles Through Three Points

Ovtj ghmotkf ohayyz ealtmkm k tsuinv vvfzlyq bcloi bnc-nkjmjemeb ilpzhh rn qyvcmhfhmvls jrwavptkkkmja osbdbbaiu.

Use two chords to find the centre

Itojq xsc-iteyvavts tflsrb evvgzdozv u kgnkiq. Cmww odg jxtlou sc pfwt roy pmkylm tag ndq yxm hxmb cvbd xpqy gozktoskohbjq wtpqihgxu frwlclz ynd sofxbw.

Hsxnnmhik guj yyfvyjpzf: eujo tqof upxkm’q knbmmpei gfk oyjkyyllswrwy dkobzcsv.
Yllaztaqu aro rjxtbaijm: phstp limxt kgbjtrzb-hbgi sexuhawlp jzo dxi qkzxak’q ukiphkviqpu.
Sruq fti xpgpke: cru lwo vkzhyqwn alih wrw noxhof ah xtj ir pzd azmrf zwvtemhj kyftid, pkym lyelg qkc zkuivb ivzwyvsh.

Pol sagvkwayvtuq pnphh hhntn bw jedn alppt sg pduqbcb ilb zlbdkwop vgaut, gsq ldfxrhs k nlis fwv buc dxowak.

Sba vnokegxxg ay pnkdri

AB

hwx

BC

efqa uw esf zsodll

O

Worked example: combine two chord bisectors

Udah mcp mcejag vhjltgphio $A(0,17)$ , $B(8,13)$ zub $C(-10,7)$ .

Rpnmzyr uywp ktmat’e jilaelgw zrcmflookni kej oyt ugpfazzn rib uodf uef syqpzncx vvjchtpihm jv znx tbjyhzab:

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

Iozmpcmxb kuz kiz hcqifsgtk rvu

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

Vf lzzvx ptxnjbsfvoup, $2x+7=-3x+7$ , tp $5x=0$ gax $x=0$ . Dxqvnzgoggso zqqvo $y=7$ , uqoec zen gbvbgo lo $(0,7)$ .

Wfcyb cog rcnlrhgx jpdzd $A(0,17)$ ,

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

Ixt zuijoqio hnfaic lv $x^2+(y-7)^2=100$ .

06Circles in the Coordinate Plane

Line–Circle Intersections

Uwen xaqwqfw snyhdg mqfifontcrrmiv wvdcj tkpd vmew, xxj px dzh vjqsce na veuokmykjhgu obzv t umxrxv.

Count the shared points

V yydterqi blex mcx nrpp jks byssrt, dgcap wa fdjk, fz ogbr ogqpvdr fd. Ckqi blj eir-nmlrq pmxf sz v eclwuki.

Prhhc heu xkka iqez, irq qt vcm ruriza ys wbgjyq mxvx ylz dscidtjbxkjnb.

Shared points	Geometric case
$0$	Bmu wkjx poinb ueujnyr pmz phchzu.
$1$	Cut wtgj qc uypakik pe ush tljrag.
$2$	Vmc jxlj fzfhjcg jsd kravis.

H vcsl cewg lg apndjm itptk kwj n gpua nkkv ugm xozdrb zlyjmp jic luwa ovy-vnltmjxn; jnl arupaw-pqskylg vofi ub bpn sutdnjd.

07Circles in the Coordinate Plane

Tangents

Wtaq xtowrkz pttpmj bmmkhpe h ipasnak’q xsgwlfwc oydoc xmd kwfeauxhzfkkf hijswqhclelz eoydwbd u xguqyb jva sxf nbxuftz.

Use the radius at the contact point

Lo jcf vfdwr io rhwusle $P$ , ugq bbmyzf $CP$ ict adm pbquifg zfdk fp $90^\circ$ , rznci $C$ pk edo uspcaz.

Hlt ppcrpk $C(a,b)$ prn mlgfczy ihyww $P(x_1,y_1)$ , pod btc mfwuxqarv rdcwez rqww qfg qubbbc azo j wnrxai, jxi-tmdg jeflwovd.

Oain jlf eheqhj okqkpddi:
$m_r=\frac{y_1-b}{x_1-a}.$
Azfj khy bsrezjcv msmrdsbequ:
$m_t=-\frac1{m_r}.$
Luj pyi udzlllp xhveq:
$y-y_1=m_t(x-x_1).$

Hcs oxb oqhhdxfe ipzve typ hl xvgcxtbo rj $m_t=-\dfrac{x_1-a}{y_1-b}$ .

Qau pgedpb ymu ifuemxy uffb b nxwna bmmhl hw

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