Question

Hallar le ecuación general de la circunferencia conociendo el centro y el radio



Please Ayuda

Answer 1

Respuesta:

Ecuación de circunferencia: $(x - h )^2 + (y - k)^2 = r^2$

Para: C(-8, -1) , r = 4

Reemplazando:

$(x - (-8) )^2 + (y - (-1))^2 = (4)^2\\\\(x+8)^2 + (y + 1)^2 = 16\\\\x^2+2(x)(8)+8^2 + y^2+2(y)(1)+1^2 = 16\\\\x^2+16x+64+y^2+2y+1=16\\\\x^2+y^2+16x+2y+65-16 = 0\\\\x^2+y^2+16x+2y+49=0$

Para: C(2, -3/2) , r = $2\sqrt{10}$

Reemplazando:

$(x-2)^2+(y-(-\frac{3}{2})^2=(2\sqrt{10})^2\\\\(x-2)^2+(y+\frac{3}{2})^2=4*10\\\\x^2-2(x)(2)+2^2+y^2+2(y)(\frac{3}{2})+(\frac{3}{2})^2 = 40\\\\x^2-4x+4+y^2+3y+\frac{9}{4} = 40\\\\ x^2+y^2-4x+3y+4+\frac{9}{4}-40 = 0\\\\ x^2+y^2-4x+3y-\frac{135}{4} = 0$

Para : C(-11/3, 0), r = $3\sqrt{7}$

Reemplazando:

$(x-(-\frac{11}{3}))^2+(y-0)^2 = (3\sqrt{7})^2\\\\(x+\frac{11}{3})^2 + y^2 = 9*7\\\\x^2+2(x)(\frac{11}{3})+(\frac{11}{3})^2+y^2 = 63\\\\x^2+\frac{22}{3}x+\frac{121}{9}+y^2-63 = 0\\\\x^2+y^2+\frac{22}{3}x+\frac{121}{9}-63 = 0\\\\x^2+y^2+\frac{22}{3}x-\frac{446}{9} = 0$