Question 1186873
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What is the equation of the circle with center at (0, 2) and tangent to the line 3x - 4y = 12
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<pre>
All you need is to find the radius of the circle, i.e. the distance from the point (0,2) 
to the given straight line  3x - 4y - 12 = 0.


There is a remarkable formula which ideally suits for this need.


    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.


Substitute the given data  a= 3, b= -4, c= -12,  {{{x[0]}}} = 0,  {{{y[0]}}}= 2  into the formula to get the distance under the question


    {{{abs(3*0 + (-4)*2 - 12)/sqrt(3^2+4^2)}}} = {{{abs(-20)/sqrt(25)}}} = {{{20/5}}} = 4.


<U>Answer</U>.  The radius of the circle is 4 units.

         The standard equation of the circle is  {{{x^2}}} + {{{(y-2)^2}}} = {{{4^2}}}.
</pre>

Solved.