SOLUTION: Without graphing, Find the radius of a circle that is tangent to the y-axis that goes through two points (1,3), (2,4). List the formulas you use and whether you solved for an spec

The equation of a circle is



Since it's tangent to the y-axis, the center (h,k), is such that
its x-coordinate h = ±r, but since the given points are in the
first quadrant, the circle cannot go into the second quadrant and
be tangent to the y-axis.  So h = r.

so we can substitute r for h





Factor the left side as the difference of squares:











Substitute (x,y) = (1,3)


 




Solve for 2r, that's the variable I solved for.

{No need to solve for r, for that would induce denominators)



Substitute (x,y) = (2,4)


 




Solve for 4r



The system of equations to solve is



Muliply both sides of the first equation by -2
to eliminate the r-terms



Adding the two equations term by term:





k = 0;  4-k = 0
          4 = k

Using k = 0








This is the graph of the circle with center (h,k)=(5,0) and radius r=5.

  


Using k = 4







The radius cannot be negative.  So there is only one solution

Edwin

1.  Draw (mentally) the segment connecting the given points (1,3) and (2,4).

    This segment has the slope 1 = .

    The centers of the two circles lie in the perpendicular bisector to this segment.

    The perpendicular bisector goes through the middle point (1.5,3.5) and has the slope -1.



2.  Let us find the radius of the "upper" circle.

    Let "p" be the distance along the perpendicular bisector from the middle point (1.5,3.5) to the center of the "upper" circle.

    Then the center of the "upper" circle is at the point (,),  and   the radius of the upper circle is .

    Since the upper circle touches y-axis, it gives the equation for "p"

         = .

    From this equation, p = (square both sides; simplify; then apply the quadratic formula) = .

    Then the radius of the upper circle is   =  = 1,   which gives  a = 1.   (1)

    The center of this circle is the point (1,4).



3.  Now, let us find the radius of the "lower" circle.

    Let "q" be the distance along the perpendicular bisector from the middle point (1.5,3.5) to the center of the "lower" circle.

    Then the center of the "lower" circle is at the point (,),  and   the radius of the lower circle is .

    Since the lower circle touches y-axis, it gives the equation for "q"

         = .

    From this equation, q = (square both sides; simplify; then apply the quadratic formula) = .

    Then the radius of the upper circle is   =  =  = 25,   which gives  b =  = 5.   (2)

    The center of this circle is the point (5,0).


3.  The equation for the "upper" circle is 

         = 1.

    The equation for the "lower" circle is 

         = 25.