# 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.