Question 1081620
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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 specific variable in the formulas.
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The solution would be much easier to understand having a plot.
But, since you directly and explicitly ask do not graph, I will not use it.


Actually, there are <U>TWO</U> such circles as you will see later.


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

    This segment has the slope 1 = {{{(4-3)/(2-1)}}}.

    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 ({{{1.5-p/sqrt(2)}}},{{{3.5+p/sqrt(2)}}}),  and   the radius of the upper circle is {{{sqrt(p^2 + (1/sqrt(2))^2)}}}.

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

        {{{sqrt(p^2 + (1/sqrt(2))^2)}}} = {{{1.5-p/sqrt(2)}}}.

    From this equation, p = (square both sides; simplify; then apply the quadratic formula) = {{{sqrt(2)/2}}}.

    Then the radius of the upper circle is  {{{a^2}}} = {{{p^2 + (1/sqrt(2))^2}}} = 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 ({{{1.5+q/sqrt(2)}}},{{{3.5-q/sqrt(2)}}}),  and   the radius of the lower circle is {{{sqrt(q^2 + (1/sqrt(2))^2)}}}.

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

        {{{sqrt(q^2 + (1/sqrt(2))^2)}}} = {{{1.5+q/sqrt(2)}}}.

    From this equation, q = (square both sides; simplify; then apply the quadratic formula) = {{{7*sqrt(2)/2}}}.

    Then the radius of the upper circle is  {{{b^2}}} = {{{q^2 + (1/sqrt(2))^2}}} = {{{100/4}}} = 25,   which gives  b = {{{sqrt(25)}}} = 5.   (2)

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


3.  The equation for the "upper" circle is 

        {{{(x-1)^2+(y-4)^2}}} = 1.

    The equation for the "lower" circle is 

        {{{(x-5)^2+y^2}}} = 25.

</pre>

Now, finally, I use all that I got and illustrate these results in the plot below.



It is my CHECK !



{{{drawing( 420, 510, -1.5, 12.5, -8.5, 8.5, 
            grid(1),

            circle(1, 4, 1),
            circle(1, 4, 0.1),

            circle(5, 0, 5),
            circle(5, 0, 0.1)

)}}}



*** SOLVED ! ***