To determine the equation of one of the diameters of the given circle, we need to first rewrite the equation of the circle in its standard form. The given equation is:
x2+y2−6x+2y=0 We will complete the square for both the
x and
y terms.
Starting with the
x terms:
x2−6x Add and subtract
(‌)2=9 :
x2−6x+9−9=(x−3)2−9Next, for the
y terms:
y2+2y Add and subtract
(‌)2=1 :
y2+2y+1−1=(y+1)2−1 Now substitute these results back into the original equation:
x2−6x+y2+2y=0 becomes:
((x−3)2−9)+((y+1)2−1)=0Simplify it:
(x−3)2+(y+1)2−10=0This is equivalent to:
(x−3)2+(y+1)2=10 This represents a circle with center at
(3,−1) and radius
√10.
The equation of a diameter of the circle must pass through the center,
(3,−1), and the origin,
(0,0). We can find the equation of the line passing through these two points using the point-slope form.
The slope of the line passing through
(3,−1) and
(0,0) is:
‌ slope ‌=‌=‌=−‌Using the point-slope form of the line equation:
y−y1=m(x−x1)where
(x1,y1)=(3,−1) and
m=−‌, we get:
y+1=−‌(x−3) Multiply through by 3 to clear the fraction:
3(y+1)=−(x−3)Simplify the equation:
3y+3=−x+3 Rearrange to get the standard form of the line equation:
x+3y=0Therefore, the equation of one of its diameters is:
Option D
x+3y=0