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If two circles touch one another, they will not have the same centre.
| Ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον. Δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΕ ἐφαπτέσθωσαν ἀλλήλων κατὰ τὸ Γ σημεῖον: λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον. Εἰ γὰρ δυνατόν, ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ διήχθω, ὡς ἔτυχεν, ἡ ΖΕΒ. Ἐπεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΖΓ τῇ ΖΒ. πάλιν, ἐπεὶ τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΓΔΕ κύκλου, ἴση ἐστὶν ἡ ΖΓ τῇ ΖΕ. ἐδείχθη δὲ ἡ ΖΓ τῇ ΖΒ ἴση: καὶ ἡ ΖΕ ἄρα τῇ ΖΒ ἐστιν ἴση, ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ζ σημεῖον κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΕ κύκλων. Ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον: ὅπερ ἔδει δεῖξαι. | If two circles touch one another, they will not have the same centre. For let the two circles ABC, CDE touch one another at the point C; I say that they will not have the same centre. For, if possible, let it be F; let FC be joined, and let FEB be drawn through at random. Then, since the point F is the centre of the circle ABC, FC is equal to FB. Again, since the point F is the centre of the circle CDE, FC is equal to FE. But FC was proved equal to FB; therefore FE is also equal to FB, the less to the greater: which is impossible. Therefore F is not the centre of the circles ABC, CDE. |