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) (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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