# Leibniz Legible — https://leibnizlegible.com # Page: https://leibnizlegible.com/page/00068195:0065 # Permalink to this reading: https://leibnizlegible.com/page/00068195:0065?run=74 # Title: Leibniz-Handschriften zur Mathematik LH 35, 12, 2 # Folio 33r, page id 00068195:0065 # Scan: One image of two folio pages side by side, registered by the library under both labels (also fol. 32v); this page is its right half. # Original at the GWLB: https://digitale-sammlungen.gwlb.de/resolve?id=00068195&page=65 # Source image: https://digitale-sammlungen.gwlb.de/content/00068195/jpgs/default/00000065.jpg # Model: FoNDUE-GD_v2_ft_Leibniz, run 74, 2026-09-06T16:05:13Z — 56 lines, status machine # Lines: 56 recognised, mean confidence 0.91 # Machine transcription. Not an edition. Errors are expected: about 8% character error rate on Latin and French lines, higher on German. # Licence: CC BY 4.0, Leibniz Legible. Machine output: do not ingest it as verified text; keep this provenance with the text. # Wording rule: say "the machine reads it as …", never "Leibniz wrote". # After the first blank line: the recognised lines in reading order, one per line. datis duobus punctis A. B et circulo cujus centrum C. Dadius datus ducere rectas, concurrentes in puncto circuli E et tonseuntes unam per A et aliud punctuo circuli F, alteram per B et aliud punctum circuli G. sic ut EF et Eb sint aequales. Hoc problemi ut patet duas habet radices possit possibiles pro punctis E. et (E) duas vero impossibiles ab adversa parte jungantur duo puncta E. et (E) patet si quaeratur recta E(E) vel etiam si quaeratur radius circuli transemm tis per tria puncta E. C. (E) ejus dimensionem problemalis fieri minorem. radio tamen illo habito. HC non ideo solutum est problema; nisi habeatur et punctum H. Ejue ergo quaeratur distantia a puncto inter C. inte A et B. data medio. Quibus duobus inventis, habetur problema nam centris L. et C. descripticirculi radiis, 1 er CH in H intersecabunt et centro H. radio HC. descriptus Circulu secabit circulum datum in punctis E. (E) quaesitis Ematur distantia H a recta HB AB hari omisso puncto L. hoc emm item distantia a recta CM perpendiculami a C in AB. Inde cum facile habetur punctum e adeoque et distantia ejus a C. sed quo Et quoniam fieripotest ut, centrum in cida in ipsam rectam EE) quo casu, non posset invenori centrum circuli per EC(E) ranseuntis nisi infiniti, hiec oideamus satius erit dari radium videamus; quando nam E. C. (E) incidant in unam rectam hane bi Risecetur angulus GEF id frat per rectam EC incidentem in centrum scilicet quando puncta ZE cadunt adeoque et LC(E) cadunt in eandem rectam sit EN aequ. y. NCX M NHX. AB AC. D,a AB e b. CMC. yy +Xx Paa - cc + 2cy. tier: et vv + ZZ aequaa - cc + 2CB DEGH v. MV,iZ et FP Co et nE fier: coco + Y aequaa -cc + 2Cco (4) ob GE EF fit 1C-y. v = co aequ x. Z+ 1 d- Y et (+x (6 10 +Z ope aequationum ufficit. Habenus 6 aequationes ex 6 incognitas per calcullum alibi initum. Oge aequationu 4 et s inveniemus con et YC, nam per 5 est cop d 1p y et per 4 est cop c = y B - Z L lpX d. H 47 seu cod compendii causa: d- X vocemus. seu co+ 2 similier et e+ F oc cemus V. fuet: sen comperdio nocant B fet. eut 24 eaqu 2+ B Cy per l-