Question 1191477
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Find the equation of the concentric circle to the circle x^2+y^2-4x+6y-17=0 which has a tangent of 3x-4y+7=0.
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By completing the square method, find the center of the circle.  It is the point (2,-3).


To find the radius of the required circle, find the distance from the center (2,-3) to the given line 3x-4y+7=0.


For finding the distance from a given point to a straight line, there is a remarkable formula.


    Let the straight line in a coordinate plane is defined in terms of its linear equation 

         a*x + b*y + c = 0,

    where "a", "b" and "c" are real numbers, and let P = ({{{x[0]}}},{{{y[0]}}}) be the point in the coordinate plane. 

    Then the distance from the point P to the straight line is equal to

        d = {{{abs(a*x[0] + b*y[0] + c)/sqrt(a^2 + b^2)}}}.


Regarding this formula, see the lesson
    <A HREF=https://www.algebra.com/algebra/homework/Vectors/The-distance-from-a-point-to-a-straight-line-in-a-coordinate-plane.lesson>The distance from a point to a straight line in a coordinate plane</A>
in this site.


Your straight line is 3x - 4y + 7 = 0.


Substitute the given data  a= 3, b= -4, c= 7,  {{{x[0]}}} = 2,  {{{y[0]}}}= -3  into the formula to get the distance under the question


    {{{abs(3*2 + (-4)*(-3) + 7)/sqrt(3^2+(-4)^2)}}} = {{{abs(25)/sqrt(25)}}} = {{{25/5}}} = 5.


Thus the radius of the circle should be 5 units.


Then the standard form equation of the circle is


    {{{(x-2)^2}}} + {{{(y+3)^2}}} = 25.      <U>ANSWER</U>
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Solved.