SOLUTION: derive the equation of a circle which passes through the point(-2,1), its tangent to the line 3x-2y-6=0 at the point (4,3)

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Question 1099107: derive the equation of a circle which passes through the point(-2,1), its tangent to the line 3x-2y-6=0 at the point (4,3)
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
derive the equation of a circle which passes through the point A(-2,1), its tangent to the line 3x-2y-6=0 at the point B(4,3)
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Step 1, find the equation of the line perpendicular to the given line thru the point (4,3) --- point B
The center is on that line.
Step 2, find the perpendicular bisector of AB.
The center is on that line, too.
The center is the intersection of the 2 lines.
r is the distance from the center to either point.
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email via the TY note for help or to check your work.

Answer by Edwin McCravy(20083)   (Show Source): You can put this solution on YOUR website!
Here's a better way.



We set the lengths of the two green radii equal:



That simplifies to:



That's one equation in h and k.

To get the other equation in h and k, we use the fact that a 
tangent to a circle is perpendicular to the radius drawn to the 
point of tangency.

We find the slope of the tangent line:






Comparing that to y = mx+b, the tangent line has slope
 so the radius drawn to the point of tangency
has slope , its "negative reciprocal".

So we use the slope formula and set its slope equal to 



That simplifies to:



So we solve this system:



Solve that by substitution or elimination and get



Then we find the length of the radius, using the distance
formula for the length of the radius drawn to the point 
of tangency:




which simplifies to 

Substituting for the center and radius in





which simplifies to:



Edwin

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