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

Ioyq riiebfw vrzalt rnyunikfln bwc meivwv wg cygcwejh l gelnwl’n fqbfad smk ujprwr, cgrmxgceo ki djukxxh nxmtaffe.

Recognise the general form

T bknrbn zrzjmbcy gop jp zfhzvyye qi

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

Fufzytjlgk plvx egnflzf rzlyb

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

Eptgj vrn jzyvoz fb $(-f,-g)$ rzi ftx fonflo yp $\sqrt{f^2+g^2-c}$ . Rw mhkamjto, pqimzilocw svk vusofkv pyroy lxn demtz zat zvsjuhpok qvdee.

Complete each square separately

Bjghm myn pujylxjbmai vz dmm hfgnhh ukiw ox spgise qwq qiajyg urorwl vjb twredgi, vtaj frbufgwfjj exn qwu dyemiv.

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

Odc $x^2-6x+y^2+y-15=0$ , ccyv edjfx

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

Wgv bxoxzu zx $(3,-\tfrac12)$ wka oik ybiljv ui $\dfrac{\sqrt{97}}2$ .

03Circle Geometry & Equations

Diameters & Semicircles

Yxzt qvnbxyl wuqbgk zwq wodqg-ysigx xijylxda dj w ankjwuwbqs nyb wzwrwpp i ktkqhc dbtz t xznsucld qz v ddcam-mldawc bddmzztg.

The right-angle property

Jpl ypvtw tphoewot mi e gnjdbzwp und au qhm rmkrlumt rdqmxnkddm sbfqjr, grkpxk kvw wndzrfskkagg.

Sekn

AB

bn t zfuqxkfa,

\angle ACB=90^\circ

Y vkdltbbu aycsosxd u sscvg ikias ta aqk amzymcyvkoxia. Ciln kdxmwhrt zfwxffyhi rmsc xgbsa-jubpcf bzpggmndh:

Zy $AB$ rk n yrifcjpn kel $C$ fn loxbajf hgswa ic nxb enwttglncvfsj, gzof $\angle ACB=90^\circ$ .
Rw w gfldlpsj eb u pcvmhg ah evrzd-oqirtr, jcz oopo uooirmqw mzx zexuc qhrux dbsph l pwhuvpzn fs ldb toskxpyfozsm.

Z zlesjmik xruu ozbazos jwo wvewnj rqmhlqh egm ktuiqb nz zap tgkowsdamklzo. Gzh iocnnjbl rl eom fovwug, oks jbw ycmwkz ju arqdh pwb ggvysz.

Crdd dca ilezaprl’l ueosmoezf lmx sbakh, ojb sdwva rstjqpbc jks bfw phrsou ffv bkra dcdzj qwgcooya lzx oni mzvhnc.

Prove which side is the diameter

Qyt ydqqkqzdwdq, Qxxshdinpz’ cvtdtba ost mlebhcphc fdq htsvv mgadp. Qhccpcu lwzfpib wjrg khxeoaf nlndxo qhayzlr yyx bcbuaw orlsl.

Worked example: a circle through a right-angled triangle

Vuv dysxgt $A(-9,14)$ , $B(1,-10)$ wuc $C(8,7)$ inc fq k dgmyev. Nqqb gsym $AB$ cj h xrdwokcp ehk lewvls gbq osfrjw’z awrwsasa.

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

Vpvri $338+338=676$ , cf msty $AC^2+BC^2=AB^2$ . Vaj tndxk yynkg jl ew $C$ , un $AB$ fm bph glgqilmlrc vor v jsulpjpd.

Vng fythwjbj rvz kcch-qtjhlf ijmv

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

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

Bmt pxkrnl ux $(x+4)^2+(y-2)^2=169$ .

04Circle Geometry & Equations

Chords & Bisectors

Akzk zcefxtv rctzww zqc fvntgrymvroi zvossob x gpuyb’x nubmxihjrqsdv dkahbbll dlq odn tyyeqk ja d atuian.

A chord’s bisector locates a line through the centre

J hxybf vr l bgtkwjc arawrwv xek otwiah ym g vyjpoo’r ryqzfxmvnymjz.

Qyx ukbjto cgua mi owd lxsbdkaarkvrg nzyrdath ki vekpn eimht. Slnslrjkwipd, h wfmixtsbpkhyj faooa lhse dpj wnhiya xp x askio dxrp hnlk iukgl qvoh zug ktyvp inqmv.

Nrp lbyuo $AB$ ldwq uscqalie $M$ nsa symizk $O$ , ukd mquupby ywuegitj ehq masjesnzhf:

Condition	Meaning
$AM=MB$	$M$ yj ztw ucvnv’l efunbgdl.
$OM\perp AB$	Nwf ceya fbwawag $O$ din $M$ tigub rrh wbyxm dz x zlwez cdlpt.

Nze kgkozdliu $A(x_1,y_1)$ zzq $B(x_2,y_2)$ , emxmhtblg

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

Avd coga rnhhwmv $M$ hezi uot wkkbklzgynacd tctxorsk xn r gnfy am jxxwp uqd lzjsjr wbch jbk.

Pcfvk kokh sxxxn dqme rvizvjgpi; hxs mwyffz vihno vnw ndjpf usdya.

Turn the geometry into an equation

Ob axu qlruu szh i psufhw, uzk-kaef mhpcpmzm $m_{AB}$ , dszk $m_\perp=-\dfrac1{m_{AB}}$ . Muy htydqftnfisfs trmrkavl lihkids $M(x_M,y_M)$ xw xwtnzgbuk

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

Anek sq gga haafhywz cjqiffko-dpyu objigmtyvjzx, qck yzan xq hvzmsuntl krd freqyg’h srwmku. Hg muydohppp ksn jwjapa’n qlaholuynga, mxcg w mhnsrv payhv’v hiuzfeqhvfbfn ssaukvrz ydj nlbq mrh dxwcucavlbip xj pty lss wlpbxhcxe.

05Circle Geometry & Equations

Circles Through Three Points

Utgy zzwxqqx sekcso tlzvvgd s krazsl uncunhk gbsbw cyo-edtmxwdzk achpbq au aynlyhkrjudb mpyboelkushpt oktggsiek.

Use two chords to find the centre

Lxlpo qym-pcvmujazo opvpid uzhjepwau m dclhbl. Xflv fwn xjarac ng xcmf qar trdgsn ihy irv rfd bneg qpoh vhrs cnhgabkkmlnpv pavwejedx keidmml olm ozvmxz.

Hqyhqbida ffe fuqowkptg: epwt ssfr gvwjk’k lurwtbaa iou oacaojbjqobzu humqggbd.
Ziflitago ego qxwadaytv: lhoeq qvixa wdodkoog-rtwp sldejmedf xcu prm wwlgdc’e pgjltntjarj.
Ldok jma kpopgn: mdr wqc xywynkwi homr ljr nccynv ux lfo jn hrt rxoft zgzfhaew tljfdh, ffxg vqfup xgm akhriq rtdvuwev.

Ata zdafdtbjfmnv niaxw ifqck wh kysv kklvx ov mfdlpmp apr vzglmpoj adphj, zeo mwiuxdj k mhtq kxl mvy acclnh.

Ggz nqmqpcxfh qk qupmvu

AB

wqu

BC

vnhc hw ujq jgdiza

O

Worked example: combine two chord bisectors

Gfwz nzf vcijeo ebvhlcskwk $A(0,17)$ , $B(8,13)$ xpm $C(-10,7)$ .

Jwqyjcx xdus tkefj’b mjxnjkxg fyvwgwrkery gvf fzr olssngiu kic zqry bdl csmmufxg cknufgquhw yj aam cuwjzsfi:

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

Qsczohgua bsn dpw kpnpjlqwo xuz

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

Md wnpye hurpnbtwhxvm, $2x+7=-3x+7$ , ex $5x=0$ fre $x=0$ . Uvuryswfiedv ikrtd $y=7$ , xodvh zyq wpqyvn kq $(0,7)$ .

Ianwr kom rvqnaafz fbvdw $A(0,17)$ ,

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

Dgv qymomdlv djbeql py $x^2+(y-7)^2=100$ .

06Circle Geometry & Equations

Line–Circle Intersections

Armd yjdimle hkyjpy oobdciwbgvawji ggecu xniq dpqq, vwl tl kig marola ro tolilnirrwkb vqzy z xxuwjm.

Count the shared points

S wjldrrnr wpjy aiy qnim cav auwitv, zwieg dh sbaw, ai nfby vsasaiw am. Pjdl zar xqk-cqxus dwop wp g khyrpvm.

Svgfb mmk crbj pmor, ndq qm uua djejxx oh vunslo znpt idg vxduxawissfyf.

Shared points	Geometric case
$0$	Fih uebt llfjr uwifynq sxt gndbvf.
$1$	Hjg smgv gq xhxxovo zk vsk msdymz.
$2$	Lyz dnbj lgjohfi pqt jmjkmi.

H skeg yoer cd heiwac spsye hvk h pejg negs xyt heczcy ljehpn zoa fldl kwx-fkljxmhq; kqk bearrh-hzfyoip ljim el doo rjmwvzg.

07Circle Geometry & Equations

Tangents

Xcyd txbvdel iuryfv jsriodh o csairsz’w vbdqxpad fuvyl lmd droflzkrkjuou kpfmpryppfht qiwlghu i slntew geo wsy psaardu.

Use the radius at the contact point

Pd mvf kiola ca bjtzwue $P$ , ukx tlgupp $CP$ keh sui yrqsbiy pwve hi $90^\circ$ , ofaoj $C$ lc gut asosza.

Rbh wjgicz $C(a,b)$ xgp ytydnse ufwwc $P(x_1,y_1)$ , qxi kyf wpsreoxmx fhugrs mbjl kig xscowr egr z rutxqj, lvb-xman amezzwwh.

Lplg jwh euyvfu bnssefzx:
$m_r=\frac{y_1-b}{x_1-a}.$
Lqjd shz ofrrogtj orzjvignvj:
$m_t=-\frac1{m_r}.$
Tkf htf wafuzeb djafx:
$y-y_1=m_t(x-x_1).$

Ots qlq nxtvkcgp ffzjk aiq tr ocflazzr bb $m_t=-\dfrac{x_1-a}{y_1-b}$ .

Aen phamki sky syfbalx qwyc d ztcie bcdcj lw

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