MHB How Do You Find the Equation of a Line Through a Point and a Circle's Center?

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To find the equation of a line through the point (3, -5) and the center of the given circle, first complete the square for the circle's equation x^2 + y^2 + 4x - 4y + 4 = 0. This will transform the equation into the standard form (x - h)² + (y - k)² = r², revealing the circle's center. With both the center and the point (3, -5) identified, calculate the slope of the line connecting these two points. Finally, apply the point-slope formula to derive the equation of the line. Completing the square is essential for determining the circle's center accurately.
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Find an equation of the line that passes through (3, -5) and through the center of the circle given.

x^2 + y^2 + 4x - 4y + 4 = 0

1. Does this question involve completing the square?

2. Must I then write the equation of the circle as

(x - h)2 + (y - k)^2 = r^2 which will disclose the center of the circle not centered at the origin. Yes?
 
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1 & 2.) Yes, complete the square on both $x$ and $y$ so that you can write the equation of the circle in the form you posted, revealing the center. Then you will have two points on the line, from which you may compute the slope, and then use the point-slope formula to get the equation of the line. (Yes)
 
MarkFL said:
1 & 2.) Yes, complete the square on both $x$ and $y$ so that you can write the equation of the circle in the form you posted, revealing the center. Then you will have two points on the line, from which you may compute the slope, and then use the point-slope formula to get the equation of the line. (Yes)

Sometimes, a little thinking goes a long way.
 
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