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Finding the Shortest Distance from a Circles Center to a Line

May 26, 2025Technology1367
What is the Shortest Distance from the Center of a Circle to a Line? T

What is the Shortest Distance from the Center of a Circle to a Line?

To find the shortest distance from the center of a circle to a given line, we first need to determine the center of the circle from its equation. In this article, we will walk through the process step-by-step, providing a comprehensive guide for readers looking to solve such problems.

Step 1: Rewrite the Circle Equation in Standard Form

Consider the equation of the circle:

(x^2 - y^2 - 6x 12y 9 0)

We can rewrite this equation by completing the square for both x and y terms.

Completing the Square

For x terms:

(x^2 - 6x (x - 3)^2 - 9)

For y terms:

(y^2 12y (y 6)^2 - 36)

Now, substitute these back into the original equation:

((x - 3)^2 - 9 (y 6)^2 - 36 9 0)

This simplifies to:

((x - 3)^2 (y 6)^2 36)

From the standard form ((x - h)^2 (y - k)^2 r^2), we can identify the center and radius of the circle:

The center of the circle is ((3, -6)) The radius of the circle is (6) since (sqrt{36} 6)

Step 2: Find the Distance from the Circle's Center to the Line

The line is given in slope-intercept form as (y 2x 10). We can rewrite it in standard form (Ax By C 0):

(2x - y - 10 0)

Here, (A 2), (B -1), and (C -10).

The formula for the distance (d) from a point ((x_0, y_0)) to the line (Ax By C 0) is:

(d frac{|Ax_0 By_0 C|}{sqrt{A^2 B^2}})

Substituting the center of the circle ((3, -6)):

(d frac{|2(3) (-1)(-6) - 10|}{sqrt{2^2 (-1)^2}} frac{|6 6 - 10|}{sqrt{4 1}} frac{2}{sqrt{5}} frac{2sqrt{5}}{5})

Final Answer: The shortest distance from the center of the circle to the line (y 2x 10) is (frac{2sqrt{5}}{5}).

Additional Methodology

Alternatively, we can use the perpendicular distance approach. The line (y 2x 10) has a slope of (2), so the perpendicular line has a slope of (-frac{1}{2}). We can find the equation of the perpendicular line passing through the center ((3, -6)).

Substitute (x 3) and (y -6) into (y -frac{1}{2}x b) to find (b):

(-6 -frac{1}{2}(3) b Rightarrow b -6 frac{3}{2} -frac{12}{2} frac{3}{2} -frac{9}{2})

The equation of the perpendicular line is:

(y -frac{1}{2}x - frac{9}{2})

Find the intersection point of this perpendicular line with (y 2x 10):

(-frac{1}{2}x - frac{9}{2} 2x 10 Rightarrow -frac{1}{2}x - 2x 10 frac{9}{2} Rightarrow -frac{5}{2}x frac{20 9}{2} frac{29}{2} Rightarrow x -frac{29}{5}) (y -frac{1}{2}left(-frac{29}{5}right) - frac{9}{2} frac{29}{10} - frac{45}{10} -frac{16}{10} -frac{8}{5})

The intersection point is ((-frac{29}{5}, -frac{8}{5})). Now, calculate the distance from the circle's center ((3, -6)) to this point:

(d sqrt{left(3 frac{29}{5}right)^2 left(-6 frac{8}{5}right)^2} sqrt{left(frac{15 29}{5}right)^2 left(frac{-30 8}{5}right)^2} sqrt{left(frac{44}{5}right)^2 left(frac{-22}{5}right)^2} sqrt{frac{1936 484}{25}} sqrt{frac{2420}{25}} sqrt{96.8} 2sqrt{5})

Final Answer: The shortest distance from the center of the circle to the line (y 2x 10) is (2sqrt{5}).

Conclusion

Both methods yield the same result, confirming the accuracy of our calculations. Whether you use the direct distance formula or the perpendicular method, the shortest distance from the circle's center to the line is (frac{2sqrt{5}}{5}).