Question 1151541
circle:

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

if a diameter whose endpoints are​ ({{{1}}},{{{3}}}) and ({{{4}}},{{{3}}}), the length is the distance between them

{{{d=sqrt(4-1)^2+(3-3)^2)}}}

{{{d=sqrt(3^2)}}}

{{{d=3}}} => {{{r=3/2}}}
so far

{{{(x-h)^2+(y-k)^2=(3/2)^2}}}


use given points to find {{{h}}} and {{{k}}}


{{{(1-h)^2+(3-k)^2=9/4}}}

{{{h^2 - 2 h + k^2 - 6 k + 10 = 9/4}}}.......eq.1


{{{(4-h)^2+(3-k)^2=9/4}}}

{{{h^2 - 8 h + k^2 - 6 k + 25 = 9/4}}}.......eq.2


subtract eq.2 from eq.1


{{{h^2 - 2 h + k^2 - 6 k + 10-(h^2 - 8 h + k^2 - 6 k + 25 ) = 9/4-9/4}}}

{{{h^2 - 2 h + k^2 - 6 k + 10-h^2 +8 h - k^2 + 6 k - 25  = 0}}}

{{{   10 +6 h  - 25  = 0}}}

{{{   6 h  - 15  = 0}}}

{{{    h  = 15/6  }}}

{{{    h  = 5/2  }}}


find k


{{{(5/2)^2 - 2 (5/2) + k^2 - 6 k + 10 = 9/4}}}.......eq.1

{{{25/4 - 5 + k^2 - 6 k + 10 = 9/4}}}

{{{  k^2 - 6 k  = 9/4-25/4+5-10}}}
{{{  k^2 - 6 k  = -9}}}

{{{  k^2 - 6 k  +9=0}}}

{{{  (k - 3)^2=0}}}=>{{{k=3}}}


and, your equation of circle is:

{{{(x-5/2)^2+(y-3)^2=(3/2)^2}}}



{{{drawing ( 600, 600, -10, 10, -10, 10, 
circle(1,3,.12), circle(4,3,.12), circle(5/2,3,.12),locate(1,3,p(1,3)),locate(4,3,p(4,3)),locate(5/2,3,C(5/2,3)),
graph( 600, 600, -10, 10, -10, 10, sqrt(-(x-5/2)^2+(3/2)^2)+3,-sqrt(-(x-5/2)^2+(3/2)^2)+3)) }}}