Question 1010983
 find the equation of circle passing through the points of intersection of the lines 
{{{x + 3y=0}}}
{{{2x- 7y = 0}}} 
--------------find intersection point
{{{x + 3y=0}}}
{{{x =-3y}}}
substitute in 
{{{2x- 7y = 0}}} 
{{{2(-3y)- 7y = 0}}} 
{{{-6y- 7y = 0}}} 
{{{-13y = 0 }}}
{{{y=0}}}

=>{{{x=0}}}

intersection point is at ({{{0}}},{{{0}}})


and whose center is point of intersection lines 
{{{x + y +1=0}}}  
{{{x-2y +4=0}}} 
----------------------------subtract
{{{x + y +1-x-(-2y)-4=0 }}}
 {{{y +1+2y-4=0}}} 
 {{{3y -3=0}}} 
{{{3y=3}}}
{{{y=1}}}

then {{{x + 1 +1=0 }}}=>{{{x +2=0}}} =>{{{x =-2}}}

so, if the center of the circle is at ({{{-2}}},{{{1}}}) =({{{h}}},{{{k}}}) and circle passes through  ({{{0}}},{{{0}}}) , its radius will be the distance between these two points 


*[invoke distance_formula 0, 0, -2, 1]

so, we have {{{h=-2}}},{{{k=1}}} and {{{r=sqrt(5)}}} and the equation of your circle is:

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

{{{(x+2)^2+(y-1)^2=5}}}



 {{{drawing( 600, 600, -5, 5, -5, 5,
circle(-2,1,.1),locate(-2,1,C(-2,1)),circle(0,0,.12),
 graph( 600, 600, -5, 5, -5, 5,-x-1,x/2+2, sqrt(-(x+2)^2+5)+1,-sqrt(-(x+2)^2+5)+1,-x/3,(2/7)x)) }}}