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(20054) About Me  (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:

%28x-r%29%5E2%2B%28y-k%29%5E2=r%5E2

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

system%28%282-r%29%5E2%2B%285-k%29%5E2=r%5E2%2C%284-r%29%5E2%2B%283-k%29%5E2=r%5E2%29

Working with the first equation:

%285-k%29%5E2=r%5E2-%282-r%29%5E2
%285-k%29%5E2=r%5E2-%284-4r%2Br%5E2%29
%285-k%29%5E2=r%5E2-4%2B4r-r%5E2
%285-k%29%5E2=-4%2B4r

Working with the second equation:

%283-k%29%5E2=r%5E2-%284-r%29%5E2
%283-k%29%5E2=r%5E2-%2816-8r%2Br%5E2%29
%283-k%29%5E2=r%5E2-16%2B8r-r%5E2
%283-k%29%5E2=-16%2B8r

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

-2%285-k%29%5E2=8-8r
%283-k%29%5E2=-16%2B8r

-2%285-k%29%5E2%2B%283-k%29%5E2=-8
-2%2825-10k%2Bk%5E2%29%2B%289-6k%2Bk%5E2%29=-8
-50%2B20k-2k%5E2%2B9-6k%2Bk%5E2=-8
-k%5E2%2B14k-33+=+0
k%5E2-14k%2B33+=+0
%28k-3%29%28k-11%29=0
k=3, k=11

Substituting k=3 in
%285-k%29%5E2=-4%2B4r
%285-3%29%5E2=-4%2B4r
2%5E2=-4%2B4r
4=-4%2B4r
8=4r
2=r

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

%28x-2%29%5E2%2B%28y-3%29%5E2=2%5E2
%28x-2%29%5E2%2B%28y-3%29%5E2=4


Substituting k=11 in
%285-k%29%5E2=-4%2B4r
%285-11%29%5E2=-4%2B4r
%28-5%29%5E2=-4%2B4r
36=-4%2B4r
40=4r
10=r

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

%28x-10%29%5E2%2B%28y-11%29%5E2=10%5E2
%28x-10%29%5E2%2B%28y-11%29%5E2=100

Edwin

Answer by solver91311(24713) About Me  (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