Question 1111770
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A circle is tangent to both the x and y-axis and the equation x+y=8. what are the equations of the circle?
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The formulation is, &nbsp;<U>OBVIOUSLY</U>, &nbsp;not exactly correct and, &nbsp;therefore, &nbsp;is not pure Math.


The correct formulation, &nbsp;simultaneously from the Math and the Grammar point of view, &nbsp;is &nbsp;<U>THIS</U>:


<pre>
    A circle is tangent to both the x and y-axis <U>and to the straight line with</U> the equation x+y=8.  Find the circles.
</pre>

<B>Solution</B>


<pre>
Since the circle is tangent to both the x and y-axis, its center lies at the bisector of the first quadrant angle  x = y,  and  r = x = y.


Now we have two cases &nbsp;&nbsp;(see the Figure below).


<U>Case 1.  Small circle inside the triangle</U>.


Coordinates of the center via the radius  x = y = r.

Coordinates of the tangent point via the radius  {{{x[t]}}} = {{{r+r/sqrt(2)}}} = {{{y[t]}}}.

Since  {{{x[t]}}} + {{{y[t]}}} = 8,  it gives an equation for "r"

    {{{r}}} + {{{r/sqrt(2)}}} + {{{r}}} + {{{r/sqrt(2)}}} = 8,

    {{{r*(1+1/sqrt(2))}}} = 4  ====>  r = {{{(4*sqrt(2))/(sqrt(2)+1)}}} = 2.34 (approx.)


{{{graph( 330, 330, -5, 20, -5, 20,
          -x + 8, 
          2.34+sqrt(2.34^2-(x-2.34)^2),  2.34-sqrt(2.34^2-(x-2.34)^2),

          13.66+sqrt(13.66^2-(x-13.66)^2),  13.66-sqrt(13.66^2-(x-13.66)^2)

)}}}


Plot &nbsp;&nbsp;x + y = 8 &nbsp;&nbsp;and two circles. 


<U>Case 2.  Large circle outside the triangle</U>.


Coordinates of the center via the radius  x = y = R.

Coordinates of the tangent point via the radius  {{{x[t]}}} = {{{R-R/sqrt(2)}}} = {{{y[t]}}}.

Since  {{{x[t]}}} + {{{y[t]}}} = 8,  it gives an equation for "R"

    {{{R}}} - {{{R/sqrt(2)}}} + {{{R}}} - {{{R/sqrt(2)}}} = 8,
 
    {{{R*(1-1/sqrt(2))}}} = 4  ====>  R = {{{(4*sqrt(2))/(sqrt(2)-1)}}} = 13.66 (approx.)


<U>Answer</U>.  Small circle radius  r = {{{(4*sqrt(2))/(sqrt(2)+1)}}} = 2.34 (approx.) and the center  x = y = r.  The equation is  {{{(x-r)^2}}} + {{{(y-r)^2}}} = {{{r^2}}}.

         Large circle radius  R = {{{(4*sqrt(2))/(sqrt(2)-1)}}} = 13.66 (approx.) and the center  x = y = R.   The equation is   {{{(x-R)^2}}} + {{{(y-R)^2}}} = {{{R^2}}}.
</pre>

Solved.