SOLUTION: Obtain the equations of the circles which touch the y axis and pass through the points (2,5) and (4,3)

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Question 1019885: Obtain the equations of the circles which touch the y axis and pass through the points (2,5) and (4,3)
Found 2 solutions by Edwin McCravy, solver91311:
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
If a circle touches the y-axis, then its radius equals
the x-coordinate of the center. That is, r=h. The center (h,k)
becomes (r,k).  

 

So the equation is of the form:



The points (2,5) and (4,3) must satisfy this equation, so we have
this system of equations to solve:



Working with the first equation:






Working with the second equation:






We can eliminate r by multiplying the results
of the first equation by -2 and adding to
the results of the second equation:










k=3, k=11

Substituting k=3 in







So the small circle has 
center (2,3) and radius 2, and it has
equation 





Substituting k=11 in







So the large circle has 
center (10,11) and radius 10, and it has
equation 




Edwin
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


Since the points and are on the desired circles, segment is a chord. Using the two-point form of a linear equation, the derived slope-intercept form of the line passing through the two given points is .

Since the center of a circle lies on the perpendicular bisector of any chord of a circle and the mid-point of segment AB is the point , the equation of the perpendicular bisector of the chord, and hence the relationship between the coordinates of either of the center points of the two desired circles is:



Since the desired circles are tangent to the -axis, the centers must lie on lines parallel to the -axis through the point of tangency. Referring to the diagram, the two centers are and , and the centers must lie on the lines and or . Also, the points of tangency are and



Note that segments , , and are all radii of the desired circles and must perforce be of equal measure. Same thing for segments , , and .

Using the distance formula without being specific about which of the centers we use gives us:




and



but we will find it convenient to use the relationship to substitute for in this distance formula.



Since the two radii must be of equal measure:



Square both sides, expand the squared binomials, and collect terms gives us:




which factors to



Hence



and



And if then , hence . Whereas if then , hence .

The standard form equation of a circle of radius centered at is .

Hence, your two equations are:



and




John

My calculator said it, I believe it, that settles it

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