Cearcall le meadhan (a,b)
Gus co-aontar cearcaill obrachadh a-mach nuair a tha fios agad air an radius agus air a' mheadhan, cleachd am foirmle: \({(x - a)^2} + {(y - b)^2} = {r^2}\), far a bheil \((a,b)\) a' riochdachadh meadhan a' chearcaill, agus far an e \(r\) an radius.
Tha an co-aontar seo an aon rud ri co-aontar coitcheann cearcaill. 'S e dìreach an dòigh sa bheil e sgrìobhte a tha eadar-dhealaichte.
Eisimpleir
Obraich a-mach co-aontar a' chearcaill leis a' mheadhan \((2, - 3)\) agus radius \(\sqrt 7\).
\({(x - 2)^2} + {(y - ( - 3))^2} = {\left( {\sqrt 7 } \right)^2}\)
\({(x - 2)^2} + {(y + 3)^2} = 7\)
Ma dh'fheumas tu e airson obair a bharrachd, faodaidh tu seo a leudachadh gus am faigh thu:
\({x^2} - 4x + 4 + {y^2} + 6y + 9 - 7 = 0\)
\({x^2} + {y^2} - 4x + 6y + 6 = 0\)
Question
Obraich a-mach co-aontar a' chearcaill le meadhan \(= (1,2)\) agus radius \(= \sqrt 5\)
\({(x - 1)^2} + {(y - 2)^2} = {\left( {\sqrt 5 } \right)^2}\)
\({x^2} - 2x + 1 + {y^2} - 4y + 4 = 5\)
\({x^2} + {y^2} - 2x - 4y = 0\)
Question
Obraich a-mach co-aontar a' chearcaill le meadhan \(= (0,0)\) agus radius \(= 4\)
\({(x - 0)^2} + {(y - 0)^2} = {\left( 4 \right)^2}\)
\({x^2} + {y^2} - 16 = 0\)
