Tusaale Su'aal Dood ah oo ku saabsan Isle'egta Goobo
Isle'egta goobada waa mowduuc muhiim ah oo ku saabsan joomatari falanqayn ah. Faham wanaagsan oo ku saabsan isle'egta goobada waa mid aad waxtar u leh, ma aha oo kaliya xisaabta laakiin sidoo kale codsiyada injineernimada iyo sayniska ee kala duwan. Maqaalkan, waxaan ka hadli doonnaa dhowr tusaale oo isle'egta goobada iyo xalalkooda. Hadafku waa in la bixiyo dulmar cad oo dhammaystiran oo ku saabsan sida loo xalliyo dhibaatooyinka ku lug leh isle'egta goobada.
Isle'egta Guud ee Goobo
Isle'egta ugu badan ee goobada ee isku-duwayaasha Cartesian waa:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Halkee:
– \( (a, b) \) waa isku-duwayaasha bartamaha goobada.
– \( r \) waa gacanka goobada.
Haddii bartamaha goobada ay ku taal barta \( (0, 0) \), isla'egta goobada waxay noqon doontaa:
\[ x^2 + y^2 = r^2 \]
Hadda, aan ka wada hadalno su'aalo tusaale ah iyo xalalkooda.
Su'aal Tusaale 1aad
Su'aal: Go'aami isle'egta goobada oo xarunteedu tahay barta (3, -2) oo leh gacan 5 ah.
Xalka:
Gunakan rumus umum persamaan lingkaran:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Ku beddel qiimayaasha \( a = 3 \), \( b = -2 \), iyo \( r = 5 \):
\[ (x – 3)^2 + (y + 2)^2 = 5^2 \]
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Haddaba, isle'egta goobada waa:
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Su'aal Tusaale 2aad
Su'aal: Go'aami isle'egta goobada oo xarunteedu tahay asalka (0, 0) oo leh gacan 7 ah.
Xalka:
Maadaama bartamaha goobada uu ku yaal asalka, waxaan isticmaali karnaa isla'egta fudud:
\[ x^2 + y^2 = r^2 \]
Ku beddel qiimaha \( r = 7 \):
\[ x^2 + y^2 = 7^2 \]
\[ x^2 + y^2 = 49 \]
Haddaba, isle'egta goobada waa:
\[ x^2 + y^2 = 49 \]
Su'aal Tusaale 3aad
Su'aal: Go'aami isle'egta goobada oo xarunteedu tahay barta (4, -5) oo taabata dhidibka Y.
Xalka:
Goobo la taaban karo oo u jeeda dhidibka Y waxay la macno tahay masaafada u dhaxaysa bartamaha goobada ilaa dhidibka Y waxay la mid tahay radiuskeeda. Masaafadani waa qiimaha buuxa ee isku-duwaha X ee xarunta goobada. Markaa, radiusku waa 4.
Gunakan rumus umum persamaan lingkaran:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Ku beddel qiimayaasha \( a = 4 \), \( b = -5 \), iyo \( r = 4 \):
\[ (x – 4)^2 + (y + 5)^2 = 4^2 \]
\[ (x – 4)^2 + (y + 5)^2 = 16 \]
Haddaba, isle'egta goobada waa:
\[ (x – 4)^2 + (y + 5)^2 = 16 \]
Su'aal Tusaale 4aad
Su'aal: Goobo waxay leedahay isle'egta \( x^2 + y^2 – 6x + 4y – 12 = 0 \). Go'aami bartamaha iyo gacanka goobada.
Xalka:
Si aan u xallino isla'egtan, waxaan u baahannahay inaan u beddelno qaab caadi ah \( (x – a)^2 + (y – b)^2 = r^2 \). Tallaabooyinka lagu dhammaystirayo waa sidan soo socota:
1. Mengelompokkan dan menyelesaikan kuadrat sempurna:
Isle'egta bilowga ah waa:
\[ x^2 + y^2 – 6x + 4y – 12 = 0 \]
Kelompokkan \( x \) dan \( y \):
\[ (x^2 – 6x) + (y^2 + 4y) = 12 \]
2. Xalli labajibbaaran ee ugu fiican:
Loogu talagalay \( x^2 – 6x \):
\[ x^2 – 6x + 9 \]
Loogu talagalay \( y^2 + 4y \):
\[ y^2 + 4y + 4 \]
Ku dar 9 iyo 4 labada dhinac ee isle'egta:
\[ (x^2 – 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 \]
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Jadi, persamaan lingkaran dalam bentuk standar adalah:
\[ (x – 3)^2 + (y + 2)^2 = 25 \]
Dari sini, kita dapat melihat pusat lingkaran adalah \( (3, -2) \) dan jari-jarinya adalah \( r = \sqrt{25} = 5 \).
Su'aal Tusaale 5aad
Su'aal: Go'aami isle'egta goobada dhex marta dhibcaha (2, 3) iyo (4, 5), oo xarunteedu tahay xariiqda x = 3.
Xalka:
Su'aasha laga qabo, waxaan ognahay in bartamaha goobada uu yahay (3, b). Goobada sidoo kale waxay dhex martaa laba dhibcood oo la yaqaan. Maadaama goobadu ay dhex marto (2, 3), masaafada u dhaxaysa bartamaha ilaa bartan waa gacanka.
Persamaan lingkaran adalah:
\[ (x – 3)^2 + (y – b)^2 = r^2 \]
Substitusi titik (2, 3):
\[ (2 – 3)^2 + (3 – b)^2 = r^2 \]
\[ 1 + (3 – b)^2 = r^2 \]
\[ (3 – b)^2 = r^2 – 1 \]
Substitusi titik (4, 5):
\[ (4 – 3)^2 + (5 – b)^2 = r^2 \]
\[ 1 + (5 – b)^2 = r^2 \]
\[ (5 – b)^2 = r^2 – 1 \]
Dari kedua persamaan, kita tahu (3 – b)^2 = (5 – b)^2. Jadi:
\[ 3 – b = \pm(5 – b) \]
Haddii \( 3 – b = 5 – b \), natiijadu run ma noqon karto. Markaa:
\[ 3 – b = -(5 – b) \]
\[ b = 4 \]
Dengan b = 4, maka persamaan lingkarannya adalah:
\[ (x – 3)^2 + (y – 4)^2 = 2 \]
Si kastaba ha ahaatee, waxaan ka xisaabin karnaa gacanka r masaafada u dhaxaysa bartamaha iyo barta (2, 3) = \(\sqrt{(2 – 3)^2 + (3 – 4)^2} \) = \(\sqrt{1+1}\) = \(\sqrt {2}\)
Isle'egta goobada waa:
\[ (x – 3)^2 + (y – 4)^2 = 2 \]
Gabagabo
Fahmidda isleegga goobada waxay fududeyn kartaa xallinta dhibaatooyin badan oo xisaabeed. Xaalad kasta, aqoonsashada bartamaha iyo gacanku waa muhiim. Waxaan rajeyneynaa, dhibaatooyinkan tusaalaha ah iyo sharraxaaddoodu waxay bixiyaan caddayn waxayna kaa caawinayaan inaad barato isleegga goobada. Ku celcelintu waxay ka dhigtaa mid ku habboon xisaabta, markaa ha ka labalabeyn inaad isku daydo dhibaatooyin kala duwan si aad u horumariso xirfadahaaga.
