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Problem 3

Find the equation of that diameter of the circle $$x^{2}+y^{2}-6 x+2 y-8=0$$ which passes through the origin.

Short Answer

Expert verified
The equation of the diameter of the given circle that passes through the origin is $$y = -1/3x$$.
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Step 1: Find the center of the circle

The standard equation of a circle is $$x^{2}+y^{2}+2gx+2fy+c=0$$. Comparing this to the given equation $$x^{2}+y^{2}-6x+2y-8=0$$, we can see that the center of the circle $$(h, k)$$ can be found by the equations $$h = -g$$ and $$k = -f$$. So $$h = -(-6/2) = 3$$ and $$k = -2/2 = -1$$ . Hence, the center of the circle is (3, -1).

Step 2: Find the slope of the diameter

A line passing through origin and a given point '(h, k) has a slope given by (k-0)/(h-0), which is simply k/h. Therefore, the slope of the diameter passing through the origin and the center of the circle (3, -1) is $$-1/3$$.

Step 3: Write the equation of the diameter

The equation of the line passing through origin with a slope m is $$y = mx$$. Here, m is $$-1/3$$, so the equation of the diameter is $$y =-1/3x$$.

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