MATHHX A
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3.6 Cirkler
Vi ved alle hvad en cirkel er, men skal vi regne på cirkler, har brug brug for en matematisk definition af en cirkel.
Læg mærke til, hvordan definitionen giver en matematisk præcis beskrivelse af det, vi intuitivt forstår som en cirkel. En cirklen er altså en samling af punkter, som har det tilfælles, at de alle har samme afstand ind til centrum. Vi
fastlægger en cirkel ved dens ligning:
-
Sætning 3.6.1
En cirkel med radius \(r\) og centrum \(C=(x_0,y_0)\) har ligningen
\[(x-x_0)^2+(y-y_0)^2=r^2\]
Lige som parablens ligning karakteriserer de punkter som ligger på parablen, karakteriserer cirklens ligningen de punkter som ligger på cirklen. Altså, de punkter \((x,y)\) som gør ligningen sand er netop de punkter som ligger på
cirklen.
-
Eksempel 3.6.1
Vi vil bestemme en ligning for cirklen med radius \(4\) og centrum i \((2,-3)\). Vi indsætter i cirklens ligning:
\[(x-2)^2+(y-(-3))^2=4^2\]
og får vores endelige ligning
\[(x-2)^2+(y+3)^2=16.\]
Læg mærke til at fortegnene inde i parenteserne er omvendte i forhold til punktets koordinater.
Øvelse 3.6.1
Opskriv ligningerne for:
-
a) cirklen med centrum i \((1,6)\) og radius \(3\).
-
b) cirklen med centrum i \((-2,2)\) og radius \(2\).
-
c) cirklen med centrum i \((0,0)\) og radius \(1\).
Løsning 3.6.1
-
a) \((x-1)^2+(y-6)^2=9\)
-
b) \((x+2)^2+(y-2)^2=4\)
-
c) \(x^2+y^2=1\)
Øvelse 3.6.2
Følgende ligninger beskriver alle cirkler. Bestem centrum og radius:
Løsning 3.6.2
-
a) Centrum er \((7,4)\) og radius er \(5\)
-
b) Centrum er \((-1,3)\) og radius er \(1\)
-
c) Centrum er \((3,0)\) og radius er \(2\)
-
Eksempel 3.6.3
Vi vil afprøve om punktet \((3,-1)\) ligger på cirklen fra eksempel 3.6.1. I eksemplet fandt vi ligningen
\[(x-2)^2+(y-3)^2=16.\]
Vi indsætter nu punktets koordinater som \(x\) og \(y\):
\[(3-2)^2+(-1-3)^2=16\]
og reducerer
\[1+16=16.\]
Det passer næsten, men vi må konkludere at punktet ikke ligger på cirklen (men tæt på).
Regner man parenteserne i cirklens ligning kommer udtrykket til at se helt anderledes ud.
-
Eksempel 3.6.4
Vi vil nu regne parenteserne i cirklens ligning fra eksempel 3.6.1.
\[(x-2)^2+(y-3)^2=16\]
Vi bruger kvadratsætningen
\[x^2+4-4x+y^2+9-6y=16\]
og samler konstanter på højreside
\[x^2-4x+y^2-6y=16-4-9\]
og reducerer
\[x^2+y^2-4x-6y=3\]
Når vi fremover støder på cirkler vil det som regel være på udfoldet form (altså uden parenteser).
-
Eksempel 3.6.5
Vi vil undersøge om ligningen \(x^2+y^2+8x-2y=-13\) kan være ligningen for en cirkel. Vi starter med at samle \(x\)’erne og \(y\)’erne
\[x^2+8x+y^2-2y=-13.\]
Nu kvadratkompletterer vi
\[(x+4)^2-16+(y-1)^2-1=-13.\]
Vi samler konstanterne på højresiden
\[(x+4)^2+(y-1)^2=-13+16+1\]
og får
\[(x+4)^2+(y-1)^2=4,\]
hvilket kan skrives som
\[(x+4)^2+(y-1)^2=2^2.\]
Vi kan altså se at vi har en cirkel med centrum i \((-4,1)\) og radius \(2\).
Øvelse 3.6.5
Omskriv til cirklens ligning:
-
a) \(x^2+y^2-8x+2y=-8\)
-
b) \(x^2+y^2-2y=0\)
Øvelse 3.6.6
Bestem centrum og radius for cirklerne med ligningerne:
Løsning 3.6.6
-
a) Centrum er \((2,-1)\) og radius er \(4\).
-
b) Centrum er \((5,0)\) og radius er \(10\).
-
Eksempel 3.6.6
Vi omskrive \(3x^2+3y^2-6x+12y=12\), så den har form som cirklens ligning. Det nye i dette eksempel er, at der står et tal foran \(x^2\) og \(y^2\) (der står 3).
Vi starter derfor med at dividerer ligningen igennem med \(3\). Nu får vi:
\[x^2+y^2-2x+4y=4,\]
og denne ligning kan vi omskrive som i Eksempel 3.6.5.
Øvelse 3.6.8
Omskriv til cirklens ligning: