MHB Finding the Equation of a Line through a Point and Circle Center?

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To find the equation of a line through the point (3, -5) and the center of the circle defined by 4x^2 + 8x + 4y^2 - 24y + 15 = 0, the circle must first be expressed in standard form by completing the square. The center of the circle is identified as the point (h, k). The slope of the line can be calculated using the coordinates of (3, -5) and (h, k). The point-slope formula can then be applied to derive the equation of the line. Understanding these steps confirms the correct approach to solving the problem.
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Find an equation of the line passing through (3, -5) and through the center of the circle 4x^2 + 8x + 4y^2 - 24y + 15 = 0.

Does this problem involve completing the square?

Must I express the above circle in the form (x - h)^2 + (y - k)^2 = r^2?

The center is the point (h, k), right?

I must then find the slope of the points (3, -5) and (h, k), right?

I then use the point-slope formula and proceed as usual.
 
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Yes, you are correct. You may use the following equation of the line passing through (x1, y1) and (x2, y2).

\[
\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}
\]

if y2 - y1 = 0, then the equation is y = y1. If x2 - x1 = 0, then the equation is x = x1.
 
Cool. Good to know that I understood the question correctly.
 
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