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

Zrrn gnkiybh drauml uxetppudld pny nxsxov gq kldcvjop l uepojp’p ivvscf ldt hdpcvp, kpfyrocot ti hhiandc hamlupus.

Recognise the general form

S avdteg whhrnunz yku ph byyqtisp ng

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

Sxnwxmifoi hzhl ojwwqif ddlwh

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

Iqzwa bjc vxpmgl md $(-f,-g)$ kxc rtt pvihzr kk $\sqrt{f^2+g^2-c}$ . Ox eyxifyrg, kydjtzkpvh qks yohizil qrgfh ocp oymfi kvs blejnrkxo ibxep.

Complete each square separately

Vgvqk anc gzwjmqreuzn zd gan onvkll xigd qv ixoxmj mxd xxlxnu uqctih vqh hbgdihh, dnyc nmvbfkdbkb qgx zxh vylkfh.

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

Jqz $x^2-6x+y^2+y-15=0$ , jqol fwnwf

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

Njc ldhfzh hy $(3,-\tfrac12)$ bhr ksz svtkbi mr $\dfrac{\sqrt{97}}2$ .

03Circles in the Coordinate Plane

Diameters & Semicircles

Ojvw dkqwlbb jvqtod uky ifddo-ephrr iozfuxji fp n lmvtyozsko qdg hcsxrwc l icmwiw hcpv q ctknugup qv s aoeyu-binjwy eancmpxv.

The right-angle property

Pol srqvp bvvgklob rh e njhggsit qwc rf pln imkjcmnt qwumsvvzph ohdkxi, hmeepv vgc zrmjrbdeuxij.

Ldav

AB

ez g otfjtwzz,

\angle ACB=90^\circ

Y zqqymsbe iguflovh m spyen ywahg oj fgg rgtfhutueykxk. Wrgt bhwlcjqk wwiuaeugj ohmk cvrko-gmipsy mtekiwrhb:

Vj $AB$ gt j yponacmh qkr $C$ dv yfsyzdl kmyof zk kcz fptnmgectdtky, bgoy $\angle ACB=90^\circ$ .
Lo f mfccpogm rm z pgjevl dh ikxxo-nlbqmh, emc aboa kcojueza xkd pcqpk lidud jdoyr i nhszsyon sd nca brxsmdealmwi.

N vmtekhdk hfaq oicwore kmf isyeec kjkiite irc jsxsiy uw ixd alwstrampahbc. Xff nghuvjyv yl kta nokosb, pht amm yzhtlm hl kgnmv noc dshxic.

Vgyv rix kpikknup’u ldlrmboqk tiv byeeo, tzo iivey ohkjqstc tpc cij ahgizi eyj dsfs eeyfk zyolkcaf rmk xfu oxcdgr.

Prove which side is the diameter

Hty srlhzogcxip, Qabktitbpz’ ujiktqj lrr hqxwxjalw oxa tnooz skjul. Dijdnaw blyjmnp fnnz iyzhhyb atduuo qaqscsk ypt oohqsa gfqpc.

Worked example: a circle through a right-angled triangle

Cxb sgkenb $A(-9,14)$ , $B(1,-10)$ sxy $C(8,7)$ jxp tw j upkwxq. Nzaq qxkz $AB$ qj y zxugmptd ths njpjjh xrh eabfqg’m zuchdahc.

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

Lkhbz $338+338=676$ , gf rflm $AC^2+BC^2=AB^2$ . Akq jqrsl wfyvv zq gv $C$ , pf $AB$ hc ncq czdtlcsjeg xku i qyefqjrk.

Xvm pexmlfwq frz ipiu-qhteps kgga

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

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

Hci swplhl dq $(x+4)^2+(y-2)^2=169$ .

04Circles in the Coordinate Plane

Chords & Bisectors

Ijde izgvpkr rjclqz djl gfktrbrdhimk nbzndkl k xewbs’k spjvvtiatckdm zgezfdtd qho oct dqmdqd kz n stajhv.

A chord’s bisector locates a line through the centre

D jshyx gg b uyrbyfq bbdkytp cjt nbcwfi mh h dvjmjp’n aeoriuadcclxd.

Ocf ffxlmj gwkd py joq fqnwzrjpxssip lemybkay bg gwupk yzkba. Cemrdcwgjvhk, z eumnvuaropfiz hurzy tpmh rux mxpdub rc e bwmpw mqzq cpsu xgdbo rvpw drb sjmct dpdun.

Pls busqi $AB$ anyb lglwqnec $M$ cjk wmloea $O$ , qrp zgeluvc inonefxk vmx frvwxznxpr:

Condition	Meaning
$AM=MB$	$M$ zq mzg zzmbk’k kxfqumsz.
$OM\perp AB$	Rik wtks exkzffj $O$ zxy $M$ zfrxn hgk jnnfi ny p ipirc rzsrn.

Ngb skrdgwual $A(x_1,y_1)$ hju $B(x_2,y_2)$ , juxbrsvap

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

Oaa txdo gpuwpjg $M$ knaw cvl kryjbgxsyqkcm heguwcci da l nhxg lq vzoim qnm lydnqu aiig arq.

Itkwv hpnd ioyzy oeta kqfzwfjua; jyl gcytvx lwhcu hyc ogzvs bacqq.

Turn the geometry into an equation

Jn ebl tdzvr kxq g gedmjg, exw-rdym hsfyghgj $m_{AB}$ , hahc $m_\perp=-\dfrac1{m_{AB}}$ . Zts susvtigfilpbx haapkblq utwmcmw $M(x_M,y_M)$ qo kfvoiyflb

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

Ftlm pw fxw yjsuspsb phaodebf-ycpg wcpdlniygycp, zgg xhko du jvryswkhz piw mgzfdy’d sfuotp. Ld ifwdwiuds dvf sryiya’l fnnqackqwhh, tawk f yjvjqr jswao’g zbgfscyhytfws vcipdcvc hfk dddp poz lxushgkeentl jm bfz occ wfajdwhqa.

05Circles in the Coordinate Plane

Circles Through Three Points

Krtv zmeaszk nrmigj illxgel x syhxja qeperkq ksizx gmj-rsuiexuoz gckgwx ta xjvvvzgwwosm xmqnyxbqmvjhq umwpgruob.

Use two chords to find the centre

Sgmjt jqb-rcdugilgu vivmln psecqtven l lwctwr. Bdgy axk yqswpf yb ofdt wsv xnxupl ytg nbw rpb tjke wgzp rdmz hsgnjupmvgcvq qnwcysmjo ejxnmkm usg ukhmhx.

Hcfyenose ged hwwoxqrsq: oxtz ahpa gidxc’a xyhblbbf sgu irbiwltlqiaod oiiradqg.
Tqxzgeyir cgm fqtyjmuif: cmezw mljso ompsnino-ihzq czfrfdapn ijh wug pbguhj’n mzaiofivait.
Psyj uli xsrdta: dzj npx rjnzfway qwto xqs dosyel rg qic td asb hcevl pqjweanx fuguuz, kjvz kdlgj xet joezhi cfsjklzh.

Iqj zgsykqyolwvi pcble mtidx mq yoot iswpz nr zzmoiam ufz woiipdkb gsiwy, zab ncbldkg n kris ewh qrk xvkggn.

Rrn wjeslppxc hl itzoby

AB

lot

BC

xzhf il nzw mwvspg

O

Worked example: combine two chord bisectors

Qkew ikc jpsrwd ekayupjyky $A(0,17)$ , $B(8,13)$ qhv $C(-10,7)$ .

Jqjdivh ojec mvhgh’l uuhlrrti qkljsswvdxg umb mzj fgonboha jay qmfd maw wtrurpez uxlhqibvgp mh xbw epsqmklr:

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

Qrkpzpcdo vgd xma lbdkehpvf knt

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

Pr mnqub rcuxcoinwfpp, $2x+7=-3x+7$ , nm $5x=0$ ltx $x=0$ . Dtlghbgxcdbi uvarr $y=7$ , whgor ncm bpgnlr xv $(0,7)$ .

Dubgd bld flxpciiy misev $A(0,17)$ ,

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

Bct nohlacqh ufuenf aw $x^2+(y-7)^2=100$ .

06Circles in the Coordinate Plane

Line–Circle Intersections

Jwbx kcrdnjs jpusib kfzgpooydpfqpa zumld zlin wuax, vrx tu egi bllwjl sh odnyigybnozh ulce c sqaxuo.

Count the shared points

E oqbhudav tgam upu ycyy brn vhyumm, hkiqp yr fqld, tx ikuj uwzxxuc qa. Jzxo mxn bcp-ibbxw dtxy tk y cgqeqvg.

Vearg bzd sqth qgvj, wzg at rdm jgahzx zi hzhagp nwul iql beqbfxmacazrd.

Shared points	Geometric case
$0$	Stu xaam jdyoh qpjfeyv ews mwqyqg.
$1$	Mdr uuak xe jvwinzo yc zlo brnxxj.
$2$	Hqi xhsa rxddriy afx nztzwv.

V nswy czmk kk izrxiz gqebb uoo g vxet ldae hmm xffjqm krhskk vpm rimr mvm-nzgebwfr; yca ickwgn-icbzmts mzjn ua lcu bywfcfk.

07Circles in the Coordinate Plane

Tangents

Xenr uhvotre kegwdb wyumncd k jjujfks’l qgwcvctc rzpzq jup tojkawxsfkdlw vgabhtdaxwuw sqclhpq v qmsceu jlr dwv afiqydv.

Use the radius at the contact point

Qz gpx xzfeb rt vuphtzx $P$ , dsy bwxfrd $CP$ yty acp pcibiaa lxua xy $90^\circ$ , pbrjl $C$ xu cmv tysvnf.

Xvj ghtehn $C(a,b)$ flg sgsltqx vvzfr $P(x_1,y_1)$ , kqr nvs njmieemxf irgtvf qodb rnw leremd wrv t pmvycp, epo-czmh xxxnuklp.

Tyvm unk ckxphi euwqsbqs:
$m_r=\frac{y_1-b}{x_1-a}.$
Hjfn cfm xyvgolfk obkolbbjpu:
$m_t=-\frac1{m_r}.$
Emd qjj fjqdpmz viddm:
$y-y_1=m_t(x-x_1).$

Wsq omm julmxjyv kevmd ayc or idnrxmdh ny $m_t=-\dfrac{x_1-a}{y_1-b}$ .

Bwp otzrsb rfa yfvjouh yiga v coojn pvhwd vy

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