Question 1130502
{{{(x-h)^2+(y-k)^2=25}}}

find intersection of line {{{2x+y=1}}} and tangent {{{3x-4y=10}}}
{{{2x+y=1}}}.......solve for {{{y}}}
{{{3x-4y=10}}}
---------------------
{{{y=1-2x}}}

substitute in {{{3x-4y=10}}}

{{{3x-4(-2x+1)=10}}}
{{{3x+8x-4=10}}}
{{{11x=14}}}
{{{x = 14/11}}}

then,
{{{y=1-2(14/11)}}}
{{{y = -17/11}}}

intersection point is:

P( {{{14/11}}},{{{ -17/11}}})


-> since center is on a line {{{2x+y=1}}} , we have

{{{2h+k=1}}}

=>{{{k = 1 - 2 h}}}

so, center C ({{{h}}},{{{k}}})=({{{h}}},{{{1 - 2 h}}})

distance from {{{P}}} and {{{C}}} is{{{ r=5}}}

{{{5=sqrt((h-14/11)^2+(1-2h+17/11)^2)}}}

{{{h = 14/11 - sqrt(5)}}} 
{{{h=-0.96334 }}} 
or 
{{{h = 14/11 + sqrt(5)}}}
{{{h=3.50879525}}}

=>{{{k = 1 - 2 (-0.96334)}}}
{{{k=2.92668 }}}
or  
{{{k=1 - 2 (3.50879525)}}}
{{{k=-6.01759}}}


so, we have two centers that lie on given line:

center C ({{{-0.96}}},{{{2.93}}} )
and
center C ({{{3.5}}},{{{-6}}})

then, we have two circles:

{{{(x+0.96)^2+(y-2.93 )^2=25}}}

or
{{{(x-3.5)^2+(y+ 6)^2=25}}}



{{{drawing ( 600, 600, -10, 10, -10, 10,
circle(-0.96,2.93,.12),circle(3.5,-6,.12),
locate(-0.96,2.93,C(-0.96,2.93)),
locate(3.50879525,-6.01759,C(3.5,-6)),locate(-4,8.5,2x+y=1),locate(7,4,3x-4y=10),
graph( 600, 600, -10, 10, -10, 10,-sqrt(25-(x-3.50879525)^2)-6.01759 ,sqrt(25-(x-3.50879525)^2)-6.01759, 3x/4-10/4,1-2x,-sqrt(25-(x+0.96334)^2)+2.92668, sqrt(25-(x+0.96334)^2)+2.92668)) }}}