Whakaoti i te x^2+y^2-6x-8y-c=0

$$x ^ { 2 } + y ^ { 2 } - 6 x - 8 y - c = 0$$

Whakaoti mō c

$$c=x^{2}-6x+y^{2}-8y$$

Ngā Upane mō te Whakaoti Whārite Rārangi

Tangohia te x^{2} mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$y^{2}-6x-8y-c=-x^{2}$$

Tangohia te y^{2} mai i ngā taha e rua.

$$-6x-8y-c=-x^{2}-y^{2}$$

Me tāpiri te 6x ki ngā taha e rua.

$$-8y-c=-x^{2}-y^{2}+6x$$

Me tāpiri te 8y ki ngā taha e rua.

$$-c=-x^{2}-y^{2}+6x+8y$$

He hanga arowhānui tō te whārite.

$$-c=8y-y^{2}+6x-x^{2}$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-c}{-1}=\frac{8y-y^{2}+6x-x^{2}}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

$$c=\frac{8y-y^{2}+6x-x^{2}}{-1}$$

Whakawehe -x^{2}-y^{2}+6x+8y ki te -1.

$$c=x^{2}-6x+y^{2}-8y$$

Whakaoti mō x (complex solution)

$$x=\sqrt{9+c+8y-y^{2}}+3$$

$$x=-\sqrt{9+c+8y-y^{2}}+3$$

Whakaoti mō x

$$x=\sqrt{9+c+8y-y^{2}}+3$$

$$x=-\sqrt{9+c+8y-y^{2}}+3\text{, }c\geq -\left(9+8y-y^{2}\right)$$