Question 1067878
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Find the equation of a circle that has radius length {sqrt(30)} and is tangent to the line 3x+y-5=0 at the point (-1,8).
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I think that more short (and more straightforward) solution is possible.


<pre>
The idea is to draw (to write the equation of) the straight line perpendicular to the given line through the given point (-1,8), 
and then to take its intersection with the circle of the radius {{{sqrt(30)}}} centered at the given point (-1,8).


{{{graph( 330, 330, -10.5, 10.5, -5.5, 15.5,
          5-3x, (x+25)/3, 8+sqrt(30 - (x+1)^2), 8-sqrt(30 - (x+1)^2)
)}}}


Plot y = 5-3x (the given line, red), y = {{{(x+25)/3}}} (the perpendicular at (-1,8), green)
and the circle centers (intersection points of the green straight line with the arcs, blue and purple arcs)



1.  The straight line perpendicular to the given line  3x+y-5=0  at the point (-1,8) is

    -(x-(-1)) + 3(y-8) = 0,   or  (which is the same)  -x - 1 + 3y - 24 = 0  or  (which is equivalent)  x = 3y-25.


2.  The circle of the radius {{{sqrt(30)}}} centered at (-1,8) is

    {{{(x+1)^2 + (y-8)^2}}} = {{{30}}}.


3.  The centers of two circles we are searching for, are the intersection points, i.e. the solutions of the system

     x = 3y-25,            (1)
     {{{(x+1)^2 + (y-8)^2}}} = {{{30}}}.    (2)


     To solve the system, substitute (1) into (2), replacing x. You will get

     {{{(3y - 24)^2 + (y-8)^2}}} = 30,

     9y^2 - 144y + 576 + y^2 - 16y + 64 = 30,

     10y^2 - 160y + 610 = 0,

     {{{y[1,2]}}} = {{{(160 +- sqrt(160^2 - 4*10*610))/(2*10)}}} = {{{(160 +- sqrt(1200))/20}}} = {{{8 +- sqrt(3))}}}.


     Thus the centers are

     a)  {{{y[1]}}} = {{{8 + sqrt(3)}}} ,  {{{x[1]}}} = 3y-25 = {{{-1 + 3*sqrt(3)}}},   and

     b)  {{{y[2]}}} = {{{8 - sqrt(3)}}} ,  {{{x[2]}}} = 3y-25 = {{{-1 - 3*sqrt(3)}}}.
</pre>

Having the centers and the radius r = {{{sqrt(30)}}}, everybody can write the standard equations for the two circles.