write the equation of the circle x^{2}+y^{2}-6x+8y=0 in the standard form. Find the radius of the circle and the distance from the center to the origin
equation of a circle
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2 Answers
x^{2 }+ y^{2 }- 6x + 8y = 0
rearrange the equation:
x^{2 }- 6x + y^{2} + 8y = 0
we use completing the square:
(x - 3)^{2} + (y + 4)^{2} = 9 + 16 = 25
so, radius is √25 or 5
also, distance from center to origin is also 5
(a ± b)^{2} = a^{2} + 2ab + b^{2}
Standard form for the equation of the circle is
(x -a)^{2} + (y - b)^{2} = r^{2}
Distance between two poins on the coordinate plane is:
d = √[(x_{1} - x_{2})^{2} + (y_{1}- y_{2})^{2}]
~~~~~~~~~~~
x^{2} + y^{2} - 6x + 8y = 0
(x^{2} - 2*3*x + 9) + (y^{2} + 2*4*y + 16) = 9 + 16
(x - 3)^{2} + (y + 4)^{2} = 5^{2}
and coordinates of the central are (3, - 4) and coordinates of origin are (0, 0), the
distance from the central to the origin is:
d = √(x^{2} + y^{2}) = √25 = 5
Standard form for the equation of the circle is
(x -a)^{2} + (y - b)^{2} = r^{2}
Distance between two poins on the coordinate plane is:
d = √[(x_{1} - x_{2})^{2} + (y_{1}- y_{2})^{2}]
~~~~~~~~~~~
x^{2} + y^{2} - 6x + 8y = 0
(x^{2} - 2*3*x + 9) + (y^{2} + 2*4*y + 16) = 9 + 16
(x - 3)^{2} + (y + 4)^{2} = 5^{2}
and coordinates of the central are (3, - 4) and coordinates of origin are (0, 0), the
distance from the central to the origin is:
d = √(x^{2} + y^{2}) = √25 = 5