Question 1191931
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Find the equation of a circle tangent to the line 3x - 4y = 34 at the point (10, -1) and also
tangent to the line 4x + 3y = 12 at the point (3, 0).
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            This problem is NOT an ORDINARY problem : it is a SPECIAL.


            It is a special, since it is OVER-DEFINED.


            It is clear that the center of the circle lies at the intersection of perpendicular lines to the given lines 
            at the given points;


            but when we will find this intersection, we should CHECK that the distance is the same from the intersection
            point to both given points.


            So I came to assist you in making all necessary steps.



<pre>
(1)  The line perpendicular to  3x - 4y = 34  is  4x + 3y = const  (you swap x and y and change the sign in the left side at one term).

     In order for the point (10,-1) would belong to this perpendicular line, the const must be  const = 4*10 + 3*(-1) = 40 - 3 = 37.

     So, first perpendicular line is  4x + 3y = 37.



(2)  The line perpendicular to  4x + 3y = 12  is  3x - 4y = const  (you swap x and y and change the sign in the left side at one term).

     In order for the point (3,0) would belong to this perpendicular line, the const must be  const = 3*3 - 4*0 = 9 - 0 = 9.

     So, second perpendicular line is  3x - 4y = 9.



(3)  To find the intersection point, you should solve this system of two equations

         4x + 3y = 37    (1)

         3x - 4y =  9    (2)


     The intersection point is (by the determinant method)

        x = {{{(37*(-4) - 9*3)/(4*(-4) - 3*3)}}} = 7;  y = {{{(4*9 - 3*37)/(4*(-4) - 3*3)}}} = 3.



(4)  So, the intersection point and the center of the circle is the point (x,y) = (7,3).


     The distance to the point (10,-1) is  {{{sqrt((10-7)^2 + ((-1)-3)^2)}}} = {{{sqrt(3^2+4^2)}}} = {{{sqrt(25)}}} = 5.

     The distance to the point  (3,0)  is  {{{sqrt((3-7)^2 + (0-3)^2)}}} = {{{sqrt((-4)^2+(-3)^2)}}} = {{{sqrt(25)}}} = 5.



(5)  The distances are the same, so the problem is posed correctly.



(6)  The standard form equation of the circle is  {{{(x-7)^2}}} + {{{(y-3)^2}}} = {{{5^2}}}.
</pre>

Solved.