Co-aontar cearcaill
Mar as trice, bidh co-aontar coitcheann cearcaill a' nochdadh san riochd \({x^2} + {y^2} + 2gx + 2fy + c = 0\)
far an e \(( - g, - f)\) meadhan a' chearcaill
agus \(\sqrt {{g^2} + {f^2} - c}\) an radius.
Cuimhnich gum feum \({g^2} + {f^2} - c\textgreater0\) airson an cearcall a bhith ann.
Thoir sùil air na h-eisimpleirean seo a th' air an obrachadh a-mach.
Eisimpleir 1
Airson \({x^2} + {y^2} + 6x - 8y - 11 = 0\)
\({g^2} + {f^2} - c = {(3)^2} + {( - 4)^2} - ( - 11) = 36\)
Mar sin tha an co-aontar a' riochdachadh cearcall le meadhan \(( - 3,4)\) agus radius \(\sqrt {36} = 6\)
Eisimpleir 2
Airson \({x^2} + {y^2} - 2x + 4y + 11 = 0\)
\({g^2} + {f^2} - c = {( - 1)^2} + {(2)^2} - 11 = - 6\)
Mar sin chan eil \({x^2} + {y^2} - 2x + 4y + 11 = 0\) a' riochdachadh cearcall.
Eisimpleir 3
Airson \(3{x^2} + 3{y^2} - 6x + y - 9 = 0\) feumaidh sinn seo a sgrìobhadh a' tòiseachadh le \({x^2} + {y^2}\):
\({x^2} + {y^2} - 2x + \frac{1}{3}y - 3 = 0\)
\({g^2} + {f^2} - c = {( - 1)^2} + {\left( {\frac{1}{6}} \right)^2} - ( - 3)\)
\(= 4\frac{1}{{36}} = \frac{{145}}{{36}}\)
Mar sin tha an co-aontar a' riochdachadh cearcall le meadhan\(\left( {1, - \frac{1}{6}} \right)\) agus radius \(\sqrt {\frac{{145}}{{36}}} = \frac{{\sqrt {145} }}{6}\)
Ceistean
Thoir sùil air na co-aontaran a leanas agus inns am b' urrainn dhaibh cearcall a riochdachadh, agus nam b' urrainn, sgrìobh an radius agus am meadhan.
Question
\(2{x^2} + 2{y^2} + 4x - 3y - 6 = 0\)
\({x^2} + {y^2} + 2x - \frac{3}{2}y - 3 = 0\)
\({g^2} + {f^2} - c = {1^2} + {\left( { - \frac{3}{4}} \right)^2} - ( - 3) = \frac{{73}}{{16}}\)
Mar sin tha an co-aontar a' riochdachadh cearcall le meadhan \(\left( { - 1,\frac{3}{4}} \right)\) agus radius \(\frac{{\sqrt {73} }}{4}\)
Question
\({x^2} + {y^2} + 2x - 4y + 6 = 0\)
\({g^2} + {f^2} - c = {(1)^2} + {\left( { - 2} \right)^2} - 6 = - 1\)
Mar sin chan eil an co-aontar a' riochdachadh cearcall.
