Question 1067878
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There will be two solutions because the circle could be on either 
side of the line.

The equation of a circle is:

(x-h)² + (y-k)² = r² with center (h,k) and radius r.

The radius is the distance from (h,k) to (-1,8).

We use the distance formula:

{{{d}}}{{{""=""}}}{{{sqrt((x[2]-x[1])^2+(y[2]-y[1])^2)}}}

{{{r}}}{{{""=""}}}{{{sqrt((h-(-1)^"")^2+(k-8)^2)}}}

{{{sqrt(30)}}}{{{""=""}}}{{{sqrt((h+1)^2+(k-8)^2)}}}

Square both sides:

{{{30}}}{{{""=""}}}{{{(h+1)^2+(k-8)^2}}}

The radius {{{sqrt(30)}}} is also the perpendicular distance from
(h,k) to the line 3x+y-5=0

We use the formula:

The perpendicular distance from the point (x<sub>1</sub>,y<sub>1</sub>)
to the line Ax+By+C=0 is d = {{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

{{{sqrt(30)}}} = {{{abs(3h+k-5)/sqrt(3^2+1^2)}}}

{{{sqrt(30)}}} = {{{abs(3h+k-5)/sqrt(10)}}}

{{{sqrt(30)sqrt(10)}}} = {{{abs(3h+k-5)}}}

{{{sqrt(300)}}} = {{{abs(3h+k-5)}}}

{{{sqrt(100*3)}}} = {{{abs(3h+k-5)}}}

{{{10sqrt(3)}}} = {{{abs(3h+k-5)}}}

{{{"" +- 10sqrt(3)}}} = {{{3h+k-5}}}

{{{3h+k-5}}}{{{""=""}}}{{{"" +- 10sqrt(3)}}}

{{{k}}}{{{""=""}}}{{{5 +- 10sqrt(3)-3h}}}

Substitute in

{{{30}}}{{{""=""}}}{{{(h+1)^2+(k-8)^2}}}

{{{30}}}{{{""=""}}}{{{(h+1)^2+(5 +- 10sqrt(3)-3h-8)^2}}}

There is a lot of work here!  I'm not going to type the
steps.  But you square those out and use the quadratic
formula.

Use the + and solve for h and get {{{h = 3sqrt(3)-1}}}
then
{{{h = -1+3sqrt(3)}}}
Then substitute that in:

{{{k}}}{{{""=""}}}{{{5 + 10sqrt(3)-3h}}}
{{{k}}}{{{""=""}}}{{{5 + 10sqrt(3)-3(-1+3sqrt(3))}}}
{{{k}}}{{{""=""}}}{{{5 + 10sqrt(3)+3-9sqrt(3))}}}
{{{k}}}{{{""=""}}}{{{8 + sqrt(3)}}}

So the center of one circle is 

(h,k) = {{{(matrix(1,3, -1+3sqrt(3),",",8 + sqrt(3)))}}}



{{{drawing(320,400,-4,12,-2,18, graph(320,400,-4,12,-2,18),
locate(-3.6,8.6,"(-1,8)"), circle(-1,8,.2),
circle(3sqrt(3)-1,8+sqrt(3),sqrt(30)),locate(4.2,9.7,"(h,k)"), 
green(line(3sqrt(3)-1,8+sqrt(3),-1,8)),circle(3sqrt(3)-1,8+sqrt(3),.2),
circle(3sqrt(3)-1,8+sqrt(3),.1),
line(-11,38,11,-28) )}}}

To find the equation we substitute for h,k, and r in

{{{(x-h)^2 + (y-k)^2 = r^2}}}

{{{(x-(-1+3sqrt(3)))^2 + (y-(8 + sqrt(3)))^2 = (sqrt(30))^2}}}

{{{(x+1-3sqrt(3))^2 + (y-8 - sqrt(3))^2 = 30}}}

Using the - sign, and doing the same thing:

(h,k) = {{{(matrix(1,3,-3sqrt(3)-1,",",8 - sqrt(3)))}}}

{{{drawing(320,400,-12,4,-2,18, graph(320,400,-12,4,-2,18),
locate(-3.6,8.6,"(-1,8)"), circle(-1,8,.2),
circle(-3sqrt(3)-1,8-sqrt(3),sqrt(30)), line(-13,44,11,-28),
locate(-6.2,6.3,"(h,k)"),
green(line(-3sqrt(3)-1,8-sqrt(3),-1,8)),circle(-3sqrt(3)-1,8-sqrt(3),.2),
circle(-3sqrt(3)-1,8-sqrt(3),.1)

 )}}}

You can get the equation of that circle the same way.

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