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.