Persamaan Lingkaran

(x + 4)² + (y - 3)² = 16

(x - 4)² + (y + 3)² = 36

(x + 4)² + (y - 3)² = 25

(x - 4)² + (y - 5)² = 25

(x + 5)² + (y - 5)² = 16

$$\left(x-8\right)^2+\left(y-3\right)^2=\sqrt[]{6}$$

$$\left(x-8\right)^2+\left(y-3\right)^2=6$$

$$\left(x+8\right)^2+\left(y+3\right)^2=\sqrt[]{6}$$

$$\left(x-8\right)^2+\left(y-3\right)^2=36$$

$$\left(x+8\right)^2+\left(y+3\right)^2=6$$

$$\left(x+2\right)^2+\left(y-3\right)^2=13$$

$$\left(x+2\right)^2+\left(y-3\right)^2=\sqrt[]{13}$$

$$\left(x-2\right)^2+\left(y+3\right)^2=13$$

$$\left(x-2\right)^2+\left(y-3\right)^2=\sqrt[]{13}$$

$$\left(x+2\right)^2+\left(y+3\right)^2=13$$

$$x^2+y^2+8x-12y-12=0$$

$$x^2+y^2-8x+12y+12=0$$

$$x^2+y^2+4x-6y-12=0$$

$$x^2+y^2-4x+6y+12=0$$

$$x^2+y^2+8x-12y-84=0$$

$$\left(1,1\right)$$ dan $$2\ \sqrt[]{3}$$

$$\left(1,1\right)$$ dan $$6$$

$$\left(0,0\right)$$ dan $$2\ \sqrt[]{3}$$

$$\left(0,0\right)$$ dan $$6$$

$$\left(0,0\right)$$ dan $$12$$

$$\left(5,-7\right)$$ dan $$32$$

$$\left(5,-7\right)$$ dan $$8$$

$$\left(-5,7\right)$$ dan $$32$$

$$\left(-5,7\right)$$ dan $$12$$

$$\left(-5,7\right)$$ dan $$8$$

$$\left(-4,3\right)$$ dan $$7$$

$$\left(-3,4\right)$$ dan $$7$$

$$\left(-3,4\right)$$ dan $$6$$

$$\left(3,-4\right)$$ dan $$7$$

$$\left(6,-8\right)$$ dan $$5$$