Imibuzo Eyisibonelo Exoxa Ngesikhundla Sephuzu Ngokuphathelene Nendilinga
Ukunquma indawo yephuzu elihlobene nendilinga kuyisihloko esibalulekile ku-geometry eyisisekelo, ikakhulukazi ekufundeni imibuthano. Kulesi sihloko, sizoxoxa ngezinkinga eziningana zezibonelo ezihilela indawo yephuzu elihlobene nendilinga, kanye nezincazelo zazo. Lokhu kuzosiza ekucaciseni umqondo ngokusebenzisa ukusetshenziswa okungokoqobo.
I-Pendahuluan
Ngaphambi kokucwila emibuzweni eyisibonelo, ake sikhumbule izikhundla ezintathu ezingaba khona zephuzu embuthanweni:
1. Ngaphakathi kwendilinga: Uma ibanga lephuzu eliya enkabeni yendilinga lincane kune-radius yendilinga.
2. Ngaphandle kwendilinga: Uma ibanga lephuzu eliya enkabeni yendilinga likhulu kune-radius yendilinga.
3. Endingilizini: Uma ibanga elisuka endaweni eya enkabeni yendingilizi lifana nerediyasi yendingilizi.
Ngokwezibalo, indawo yephuzu \(T(x_1, y_1)\) maqondana nendilinga ephakathi ku \((a, b)\) enerediyasi \(r\) inganqunywa ngokuqhathanisa \(T(x_1, y_1)\) nesibalo sendilinga, okungukuthi:
\[
(x – a)^2 + (y – b)^2 = r^2
\]
Uma umphumela wokufaka esikhundleni se-\(x_1\) kanye ne-\(y_1\) ku-equation unikeza inani:
– Incane kune-\(r^2\), iphuzu lingaphakathi kwendilinga.
– Kukhulu kuno-\(r^2\), iphuzu lingaphandle kwendilinga.
– Kufana nokuthi \(r^2\), iphuzu lisendilinga.
Imibuzo Eyisibonelo Nengxoxo
umbuzo 1
Nquma indawo yephuzu \(T(3, 4)\) maqondana nendilinga ene-equation \( (x – 1)^2 + (y – 2)^2 = 25 \).
Ingxoxo:
Isinyathelo sokuqala ukuhlola i-equation yendilinga bese uthola ibanga ukusuka ephuzwini \(T(3, 4)\) kuya enkabeni yendilinga \((1, 2)\).
1. Khomba isikhungo kanye nerediyasi yendilinga:
Isibalo sendilinga: \( (x – 1)^2 + (y – 2)^2 = 25 \)
– Isikhungo sendilinga (\(a, b\)): (1, 2)
– Irediyasi yesiyingi (\(r\)): \(\sqrt{25} = 5\)
2. Bala ibanga eliphakathi kwephuzu \(T(3, 4)\) kanye nesikhungo sendilinga \( (1, 2) \):
\[
D = \sqrt{(3 – 1)^2 + (4 – 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
Inani \( 2\sqrt{2} \cishe 2 \izikhathi 1.414 = 2.828 \) (ngaphansi kuka \(5\)).
3. Isiphetho:
Njengoba \( 2\sqrt{2} < 5 \), khona-ke iphuzu \( T(3, 4) \) lingaphakathi kwendilinga. Umbuzo 2 Indilinga inesikhungo endaweni \( (0, 0) \) kanye nerediyasi engu-7. Thola indawo yephuzu \(P(5, 6)\) maqondana nendilinga.
umbuzo 3
Nquma indawo yephuzu \(M(2, -1)\) maqondana nendilinga ene-equation \( x^2 + y^2 = 5 \).
Ingxoxo:
1. Bala ibanga ukusuka endaweni \(M(2, -1)\) kuya enkabeni yendilinga \( (0, 0):
\[
D = \sqrt{(2 – 0)^2 + (-1 – 0)^2} = \sqrt{4 + 1} = \sqrt{5}
\]
2. Qhathanisa ibanga \(D\) nerediyasi yendilinga:
Irediyasi yesiyingi (\(r\)) = \(\sqrt{5}\).
3. Isiphetho:
Njengoba \( \sqrt{5} = \sqrt{5} \), khona-ke iphuzu \( M(2, -1) \) lisendilinga.
umbuzo 4
Indilinga enesikhungo ku-\( (4, 3) \) inobubanzi \(\sqrt{10}\). Khombisa indawo yephuzu \( N(7, 7) \) maqondana nale ndilinga.
Ingxoxo:
1. Isilinganiso Sendilinga:
Isibalo sendilinga enesikhungo \( (4, 3) \) kanye nerediyasi \( \sqrt{10} \) sithi:
\[
(x – 4)^2 + (y – 3)^2 = 10
\]
2. Bala ibanga ukusuka endaweni \( N(7, 7) \) kuya enkabeni yendilinga \( (4, 3) \):
\[
D = \sqrt{(7 – 4)^2 + (7 – 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
3. Isiphetho:
Njengoba \( 5 > \sqrt{10} \), khona-ke iphuzu \( N(7, 7) \) lingaphandle kwendilinga.
I-Penutup
Ngokuqonda ukuthi singabala kanjani ibanga lephuzu ukusuka enkabeni yendilinga bese siliqhathanisa ne-radius, singanquma kalula indawo yephuzu maqondana nendilinga. Ingxoxo kulesi sihloko kulindeleke ukuthi inikeze ukuqonda okucacile komqondo kanye nendlela yokuxazulula izinkinga ezihilela indawo yephuzu maqondana nendilinga.
Empeleni, ukwazi indawo yala maphuzu kuwusizo kakhulu ekusetshenzisweni kwezibalo okuhlukahlukene, okuhlanganisa ukuhlaziywa kwejiyomethri, ukwakheka kwezithombe, kanye nobunjiniyela. Ngakho-ke, ukuqonda lo mqondo kuyisisekelo esibalulekile esidinga ukunakwa ngokucophelela nokuqonda okujulile.
