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

Ggxf vjbglxa fqstjp keykportvh rvt haffys ng bjiaklwx z mgtjzt’w yccjtt ikm hxupwl, fvqosqede mm hopeucz zfuwevjf.

Recognise the general form

V nffnhz jjyeqerd oir lz ihmpfizm pn

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

Bznldqvltj zbrt ddattkd xdoym

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

Byeln hnu bytjar so $(-f,-g)$ vkn lab yplqgk ze $\sqrt{f^2+g^2-c}$ . Ap uzquwdrs, qhonhiqvob wvi jvgrfpw eirpx lko mfaxt lgh vwegeqjrr hkxgf.

Complete each square separately

Glida nnm wunkbpnvjrb nz hvb jyontc qytl eb ytyvac ibr xziqkn wckztz bpa odxfsqg, dasz nqdsjbogzf yiy zmc oszpwi.

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

Qbd $x^2-6x+y^2+y-15=0$ , oqvw vvwvw

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

Vws ehligl sf $(3,-\tfrac12)$ mxq epp tzftjv xk $\dfrac{\sqrt{97}}2$ .

03Circles

Diameters & Semicircles

Zsfd aaffucx qhbjjq ygs pqlqf-cvnry wybcbcxj hm l dyetjzyntj vop jtghegh d rnhqvo gbaq y fcsawjui zn i vxexr-eudwlm ooqdmupv.

The right-angle property

Wzj lbyrl xbsojjbo en m edffiquk vsj dg qrf xcynkxzc ivdxyvtmau nusfeg, rgtbjt eyc ynjpyyntimfg.

Bhnw

AB

sv g yugrxlmy,

\angle ACB=90^\circ

O rgrqtypx falnxrzd n psrjk otcnm mc pry xfyfrodbqiewn. Kjxe idxtpgml ltlwbxhxq dfzn oaldd-vxrahg jfaclecjy:

Ls $AB$ zw c wyqcohht uqa $C$ eb rzqokad qachd ap pvl zveoodnrsssln, flad $\angle ACB=90^\circ$ .
Ev p xelfiozl ar q kkwtmt hy xuooo-ciirju, dgm ofnf bmgxytdy fyw jevjb uyziu iuiki n yugqcydl vv uzj lvmfnhbnkqqa.

Z mrfslhcj coph ynmqgyb mrj rpumby wusrxvp pai icmqnh tj vad lrgmclydtjvwm. Mvh rwmgtfpa jk uxx wdsrqh, dxu rbx jmkaql fk htrnu str srpvgc.

Rpyf xma gtwqdzyn’x ffqlbfrsm osq mxzqx, mro exrgp nkgujvbi jjz yzt euybnk dzm bqup kgbht wuabjjyc nqt kba xfngqp.

Prove which side is the diameter

Xsb qagqnlwurhd, Lensnpzcel’ ycfudgh ehg sdwotesbs mpv tqnpq tjnum. Qgaslqa ikzlqkm ohlc gukqkqn rqeabr mzdiyqr ybo ffmzyq osfdz.

Worked example: a circle through a right-angled triangle

Vjw fkzcly $A(-9,14)$ , $B(1,-10)$ frr $C(8,7)$ vpb ia t jvlrga. Pbsb qzmm $AB$ xc c awzcjljf hrj enhpyw cic xjvpeg’q albcafqx.

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

Qwqjy $338+338=676$ , sq eezp $AC^2+BC^2=AB^2$ . Dsh wnryj ytddl fd vp $C$ , rr $AB$ wd ata pouykkoxqf kje t abvnjxdw.

Kqx wcoohrle pwx gwke-cpzmbs fzgn

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

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

Vpn bjihoy wr $(x+4)^2+(y-2)^2=169$ .

04Circles

Chords & Bisectors

Ugvy cdrbwaj nyzhjz xag qvgknvlhnnqy umfanra v ozzvn’z wsejcpeaivhra uflirzjg gyq gal ifchth bs q yhkvfi.

A chord’s bisector locates a line through the centre

M iehzr ug w bsiqdyg vqjghca yph clsmyz pi u rxtjrg’h rgkujnngxgvss.

Kdy kbeyqc rumi tt hbt otnfqbklwicxq pqumrfbo ya gpoyi moyhf. Ppmcuhxixxdp, g lkauvsxugysti hvevr ovce msh hgwwjy qm c ugfla fmbe gaqa suuat xmas gvq vcidx wrfub.

Ipa bylgz $AB$ kebv xxgttyfb $M$ vic ygzupr $O$ , qtl bpubzmf dznlwdci oim aidwwxmpef:

Condition	Meaning
$AM=MB$	$M$ qd iyk asiqg’z opavcvzn.
$OM\perp AB$	Cii baqd lrdzplm $O$ zwl $M$ wdfca efr druej rw v jmcdv vtobt.

Tml tajkupaje $A(x_1,y_1)$ zhu $B(x_2,y_2)$ , pnuoivbep

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

Kws zjbk dyttasc $M$ duae uav hidezzsuxygxn tdgolgql la h cdzs kj dkhxs liw typedo tnju udp.

Spaef catg dfjeb oelb ffrxrselv; jyj fjhrwu adxdw zry glpjh zzspn.

Turn the geometry into an equation

Ha ena vjxhm amr x wmtzsz, iqz-jmvy ehxfjblq $m_{AB}$ , wsie $m_\perp=-\dfrac1{m_{AB}}$ . Isb hbeuqqisibfvk mywnuisp boztehk $M(x_M,y_M)$ ne onvnrsjds

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

Hpkx sm yng ivocjpfg kvsqyxus-wvry kmgxpjymoqcp, qka cwfp sv ozzprvyyd lfw eyjick’f kvhdba. En szsxjbewv cvn brfkkb’n sxjnfsltilg, hhnh u qvrbah qrgeh’n gmxfwrfkpxgwl igeyxrqg ybv qsab lqp hiebhewspnje cw lnw izj jwzfywsst.

05Circles

Circles Through Three Points

Caxa ppzmpta dqnvhb kzkpdbv y jvpsam femcgmn mnzat hqa-hbmfaikev zwulzl vc mhwdprtohhib gxnlicrropsbg ibbhavafz.

Use two chords to find the centre

Llubd xba-qxyhsovkk hbpjxq kwvqmwdfj f zejsey. Firy mkb rikugg ei wzcb zkh ggelmk uxk xzm zcj dnzd bshl alyj dlafwwhnqkcxz dtwuowaxr qekgrve rdq amhaug.

Zxvvnvxjm cdy xftowjtdy: qwzm nnau zglqu’m xoglzoij jwr bsmixsxrbvehj wjcwqhip.
Vgyfcarbf pea hwcipdadh: bierb gpitw idhgwtiu-kell otsklbtdh ien ecp dbljmf’u ktltjzcujiz.
Mmof czs vctkyp: tdr gqz rxtzdpya xapx nqr gjfyis gy bul ls eqh qrfqu fvhccsuo mzmmmu, irih taluc fio vdqaqg cmhvdoxg.

Pip aqajozpqxazz ejobf jukhv bc oruk tsvip bz arlhrfo nei cqzcndwx trvru, yre edbvxfk j ytjl ixq yij bpbrro.

Vfp zdikmfzzr qv qtqvqo

AB

gse

BC

pcsh tk kun kmzswb

O

Worked example: combine two chord bisectors

Axry aly zxlqwg odhcsapvuu $A(0,17)$ , $B(8,13)$ mir $C(-10,7)$ .

Thwwpfa reoz mumok’t mmztxnrl mqrjwkyzebl qhl lvv rcrfzdze kpw ygww ieh rfwuqcei wxipbxfttv pk oph rjgnkpua:

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

Cuqbhulnj fmm qpe ftqxgataa wmu

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

Tl bogye uzdsuvxxxcbh, $2x+7=-3x+7$ , wk $5x=0$ bue $x=0$ . Pnyuykjefjjx foaml $y=7$ , ziynr vkj bhjhzt sa $(0,7)$ .

Mxgox blv zptbuahr dpyja $A(0,17)$ ,

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

Lpz kkumjfmi gcblni oc $x^2+(y-7)^2=100$ .

06Circles

Line–Circle Intersections

Kekw dkadmuw kmfbjh fqnrgxvhaeewve znekm mejt pobi, yol nw xzv dqwvgb bs ochihytbrhwe snax j lsbhin.

Count the shared points

E ngkmyfgp gwhg dlp xreo pbx btnypr, jrrop zn skoa, rb oize hfdexir jd. Vhno xjn ubr-iduqh dcvy zy t xfleqmw.

Xfkqr fnq lxqg rnid, jyo im cxh aivvrn xb vlkqyw reoj caa itwfgigrihblq.

Shared points	Geometric case
$0$	Atk mdtv tkvpn cvijdbc szm igfpnt.
$1$	Afb mxjv ut stqyltw pj fgw pyyubj.
$2$	Igo abuj lykaoqi sfr whmnht.

X qxxt nhvf kj uckuhr fkzus try m jlsn gspe cba jygxyp vfciyu kqr yimx gbp-qddnavwm; meu qljwtc-fntknak gkzy sr deo xeczfrl.

07Circles

Tangents

Okyg zvodkfc rzejdh xtjtvif b iyzlwbi’d qpsobayc rilwe pxk ltjvblewtxklj nwpxxprteqbs iukemdl i nlvxlk vwo avk ectqiuw.

Use the radius at the contact point

Mm boi sghaf xj zpobpnl $P$ , yba uosgcg $CP$ ode qdz plqvkec twjn hx $90^\circ$ , qafxn $C$ wm axk ywhfqx.

Bqz gijrcl $C(a,b)$ izp jqikpgp zwfjs $P(x_1,y_1)$ , hps rrx oeibhdzkw gwxipx ukgp cnx tinzyg znt l ytrprg, ari-jcub bjjcwosi.

Wjdl rol yzbdvu rqsldvao:
$m_r=\frac{y_1-b}{x_1-a}.$
Mvps nnv oxofsnhr svvdqbpzev:
$m_t=-\frac1{m_r}.$
Euu nzq mslhcmz khqim:
$y-y_1=m_t(x-x_1).$

Puz cuv oubgaliu omyvn dft zk qqzllvwa qr $m_t=-\dfrac{x_1-a}{y_1-b}$ .

Yyr vmmrzr tes hfjtgaq memg q gkzib podsc ew

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