Asked by mwaba
the circle is given by the equation x^2+y^2+8x-7/2y=0
find
(i)coordinates of the centre and the radius
(ii)coordinates of the point where the circle crossesthe x-axis
find
(i)coordinates of the centre and the radius
(ii)coordinates of the point where the circle crossesthe x-axis
Answers
Answered by
Damon
oh well
x^2 + 8 x = -y^2 + 3.5 y
x^2 + 8 x + 16 = -y^2 + 3.5 y + 16
(x+4)^2 = -y^2 + 3.5 y + 16
y^2 - 3.5 y = -(x+4)^2 + 16
y^2 - 3.5 y + 3.0625 + (x+4)^2 = 19.0625
(x+4)^2 + (y-1.75)^2 = 4.366^2
x^2 + 8 x = -y^2 + 3.5 y
x^2 + 8 x + 16 = -y^2 + 3.5 y + 16
(x+4)^2 = -y^2 + 3.5 y + 16
y^2 - 3.5 y = -(x+4)^2 + 16
y^2 - 3.5 y + 3.0625 + (x+4)^2 = 19.0625
(x+4)^2 + (y-1.75)^2 = 4.366^2