Question 1142649
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<pre>
The center of this circle lies on the straight line perpendicular to the given line and passing through the point (-3,0).


From the other side, the center of this circle lies on the perpendicular bisector to the segment connecting the points (-3,0) and (-3,6).


Notice that this segment, connecting the points (-3,0) and (-3,6), is VERTICAL and its middle point is (-3,3).


So, that perpendicular bisector is simply the horizontal line y = 3.



Perpendicular to the given line has an equation

    3x + 2y = c,      (1)

where "c" is unknown constant.


Since this perpendicular line (1) passes through the point (-3,0), we have

   3*(-3) + 2*0 = c = -9.


Thus, the center of the circle lies on the line

    3x + 2y = -9


and on the line  y =3  at the same time,  which gives you

    3x + 2*3 = -9,

    3x       = -9 - 6 = -15,

     x                = -15/3 = -5.


Thus the center of the circle is the point (x,y) = (-5,3).



Since the distance from the center of the circle to the point (-3,0) is

    {{{sqrt((-5)-(-3)^2 + (3-0)^2)}}} = {{{sqrt(2^2 + 3^2)}}} = {{{sqrt(13)}}},

the radius of the circle is {{{sqrt(13)}}}.



Then the equation of the circle in the standard form is

    {{{(x-(-5))^2}}} + {{{(y-3)^2}}} = {{{(sqrt(13))^2}}},

or, equivalently,

    {{{(x+5)^2}}} + {{{(y-3)^2}}} = 13.      <U>ANSWER</U>
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If you want to see many other similar solved problems, you can find them in the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Find-the-equation-of-the-circle-given-by-its-center-and-tauching-a-given-line.lesson>Find the standard equation of a circle</A>

in this site.