Misalin Tambayar Tattaunawa kan Daidaito na Da'ira
Daidaiton da'ira muhimmin batu ne a fannin nazarin lissafi. Fahimtar daidaiton da'ira yana da matuƙar amfani, ba kawai a fannin lissafi ba, har ma a fannoni daban-daban na injiniyanci da kimiyya. A cikin wannan labarin, za mu tattauna misalai da dama na daidaiton da'ira da mafitarsu. Manufar ita ce samar da cikakken bayani game da yadda za a magance matsalolin da suka shafi daidaiton da'ira.
Daidaito na Gabaɗaya na Da'ira
Mafi yawan lissafin da'ira a cikin daidaitawar Cartesian shine:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Ina:
– \( (a, b) \) sune daidaitattun tsakiyar da'irar.
– \( r \) shine radius na da'irar.
Idan tsakiyar da'irar yana a wurin \( (0, 0) \), lissafin da'irar zai kasance:
\[ x^2 + y^2 = r^2 \]
Yanzu, bari mu tattauna wasu misalai na tambayoyi da kuma hanyoyin magance su.
Misali Tambaya ta 1
Tambaya: Kayyade lissafin da'irar da ke tsakiyarta a wurin (3, -2) kuma tana da radius na 5.
Mafita:
Yi amfani da dabarar gabaɗaya don lissafin da'ira:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Maye gurbin dabi'un \(a = 3 \), \(b = -2 \), da \(r = 5 \):
\[ (x - 3)^2 + (y + 2)^2 = 5^2 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
Don haka, daidaiton da'irar shine:
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
Misali Tambaya ta 2
Tambaya: Kayyade lissafin da'irar da ke tsakiyarta a asalin (0, 0) kuma tana da radius na 7.
Mafita:
Tunda tsakiyar da'irar tana a asalin, zamu iya amfani da lissafi mai sauƙi:
\[ x^2 + y^2 = r^2 \]
Maye gurbin ƙimar \( r = 7 \):
\[ x^2 + y^2 = 7^2 \]
\[ x^2 + y^2 = 49 \]
Don haka, daidaiton da'irar shine:
\[ x^2 + y^2 = 49 \]
Misali Tambaya ta 3
Tambaya: Kayyade lissafin da'irar da cibiyarta take a wurin (4, -5) kuma ta taɓa axis ɗin Y.
Mafita:
Da'ira mai lankwasawa zuwa ga axis ɗin Y yana nufin nisan da ke tsakanin tsakiyar da'irar zuwa ga axis ɗin Y daidai yake da radius ɗinsa. Wannan nisan shine cikakken ƙimar daidaitawar X na tsakiyar da'irar. Don haka, radius ɗin shine 4.
Yi amfani da dabarar gabaɗaya don lissafin da'ira:
\[ (x – a)^2 + (y – b)^2 = r^2 \]
Maye gurbin dabi'un \(a = 4 \), \(b = -5 \), da \(r = 4 \):
\[ (x - 4)^2 + (y + 5)^2 = 4^2 \]
\[ (x - 4)^2 + (y + 5)^2 = 16 \]
Don haka, daidaiton da'irar shine:
\[ (x - 4)^2 + (y + 5)^2 = 16 \]
Misali Tambaya ta 4
Tambaya: Da'ira tana da lissafin \( x^2 + y^2 – 6x + 4y – 12 = 0 \). Kayyade tsakiya da radius na da'irar.
Mafita:
Domin warware wannan lissafi, muna buƙatar mayar da shi zuwa tsari na yau da kullun \( (x – a)^2 + (y – b)^2 = r^2 \). Matakan kammala shi sune kamar haka:
1. Rukunin da kuma warware murabba'ai masu kyau:
Daidaito ta farko ita ce:
\[ x^2 + y^2 – 6x + 4y – 12 = 0 \]
Rukuni \( x \) da \( y \):
\[ (x^2 – 6x) + (y^2 + 4y) = 12 \]
2. Warware cikakken murabba'i:
Domin \( x^2 – 6x \):
\[ x^2 – 6x + 9 \]
Domin \( y^2 + 4y \):
\[ y^2 + 4y + 4 \]
Ƙara 9 da 4 a ɓangarorin biyu na lissafin:
\[ (x^2 – 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
Don haka, lissafin da'ira a cikin tsari na yau da kullun shine:
\[ (x - 3)^2 + (y + 2)^2 = 25 \]
Daga nan, za mu iya ganin cewa tsakiyar da'irar shine \( (3, -2) \) kuma radius shine \( r = \sqrt{25} = 5 \).
Misali Tambaya ta 5
Tambaya: Kayyade lissafin da'irar da ta ratsa ta cikin maki (2, 3) da (4, 5), kuma wacce cibiyarta take kan layin x = 3.
Mafita:
Daga tambayar, mun san tsakiyar da'irar shine (3, b). Da'irar kuma tana ratsa wurare biyu da aka sani. Tunda da'irar tana ratsawa ta (2, 3), nisan daga tsakiya zuwa wannan wurin shine radius.
Daidaiton da'ira shine:
\[ (x – 3)^2 + (y – b)^2 = r^2 \]
Ma'aunin maye gurbin (2, 3):
\[ (2 – 3)^2 + (3 – b)^2 = r^2 \]
\[ 1 + (3 – b)^2 = r^2 \]
\[ (3 – b)^2 = r^2 – 1 \]
Ma'aunin maye gurbin (4, 5):
\[ (4 – 3)^2 + (5 – b)^2 = r^2 \]
\[ 1 + (5 – b)^2 = r^2 \]
\[ (5 – b)^2 = r^2 – 1 \]
Daga lissafin guda biyu, mun sani (3 – b)^2 = (5 – b)^2. Don haka:
\[ 3 – b = \pm(5 – b) \]
Idan \( 3 – b = 5 – b \), sakamakon ba zai zama gaskiya ba. Don haka:
\[ 3 – b = -(5 – b) \]
\[ b = 4 \]
Tare da b = 4, lissafin da'irar shine:
\[ (x – 3)^2 + (y – 4)^2 = 2 \]
Duk da haka, za mu iya ƙididdige radius r daga nisan da ke tsakanin tsakiya da ma'auni (2, 3) = \(\sqrt{(2 – 3)^2 + (3 – 4)^2} \) = \(\sqrt{1+1}\) = \(\sqrt {2}\)
Daidaiton da'irar shine:
\[ (x – 3)^2 + (y – 4)^2 = 2 \]
Kammalawa
Fahimtar lissafin da'ira na iya sauƙaƙa magance matsalolin lissafi da yawa. A kowane hali, gano tsakiya da radius yana da mahimmanci. Da fatan waɗannan misalan matsalolin da bayaninsu suna ba ku haske kuma suna taimaka muku koyon lissafin da'ira. Aiki yana sa ya zama cikakke a lissafi, don haka kada ku yi jinkirin gwada matsaloli daban-daban don inganta ƙwarewar ku.
