Question 1124597
a circle is on the line {{{y=2x-3}}} and also passes through A({{{5}}},{{{2}}}) and B({{{3}}},{{{-2}}})

graph a line and given points:

{{{y=2x-3}}}

{{{x}}}|{{{y}}}
{{{0}}}|{{{-3}}}
{{{x}}}|{{{0}}}->{{{0=2x-3}}}->{{{3=2x}}}->{{{x=3/2}}}

{{{drawing(600, 600, -10, 10, -10, 10,
circle(0,-3,.12),circle(3/2,0,.12),circle(5,2,.12),circle(3,-2,.12),
locate(5,2,A(5,2)),locate(3,-2,B(3,-2)),
 graph(600, 600, -10, 10, -10, 10, 2x-3)) }}}


find coordinates of the midpoint of A({{{5}}},{{{2}}}) and B({{{3}}},{{{-2}}})

({{{(5+3)/2}}},{{{(2-2)/2}}})

({{{4}}},{{{0}}})...use it to find equation of a line perpendicular to the line {{{y=2x-3}}}-> slope is {{{m=2}}}

perpendicular lines have slopes negative reciprocal to each other

so, our perpendicular line, {{{y=mx+b}}}, will have a slope  {{{m[p]=-1/2}}}

{{{y=-(1/2)x+b}}}...to find {{{b}}} plug in coordinates of the midpoint ({{{4}}},{{{0}}})

{{{0=-(1/2)4+b}}}

{{{0=-(1/cross(2))cross(4)2+b}}}

{{{0=-2+b}}}

{{{b=2}}}

line is:

{{{y=-(1/2)x+2}}}

let's add it to the graph


{{{drawing(600, 600, -10, 10, -10, 10,
circle(0,-3,.12),circle(3/2,0,.12),circle(5,2,.12),circle(3,-2,.12),
locate(5,2,A(5,2)),locate(3,-2,B(3,-2)),
 graph(600, 600, -10, 10, -10, 10, 2x-3,-(1/2)x+2)) }}}


the point where lines {{{y=2x-3}}} and {{{y=-(1/2)x+2}}} should intersect in one point which is the center o the circle

{{{y=2x-3}}} 
{{{y=-(1/2)x+2}}}
-------------------solve system; since left sides are same, we have

{{{2x-3=-(1/2)x+2}}} ...solve or {{{x}}}


{{{2x+(1/2)x=3+2}}}

{{{2x+(1/2)x=5}}}...both sides multiply by {{{2}}}

{{{4x+x=10}}}

{{{5x=10}}}

{{{x=2}}}

now find {{{y}}}

{{{y=2*2-3}}}
{{{y=1}}}

so, intersection point is: C({{{2}}},{{{1}}})=>{{{h=2}}} and {{{k=1 }}} coordinates o the center

add to graph

{{{drawing(600, 600, -10, 10, -10, 10,
circle(0,-3,.12),circle(3/2,0,.12),circle(5,2,.12),circle(3,-2,.12),
locate(5,2,A(5,2)),locate(3,-2,B(3,-2)),circle(2,1,.12),locate(2,1,C(2,1)),
 graph(600, 600, -10, 10, -10, 10, 2x-3,-(1/2)x+2)) }}}


now we need radius {{{r}}}
since circle also passes through A({{{5}}},{{{2}}}) and B({{{3}}},{{{-2}}}), use one of these points and find distance between that point and center


C({{{2}}},{{{1}}})
A({{{5}}},{{{2}}})

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

{{{r=sqrt((5-2)^2+(2-1)^2)}}}

{{{r=sqrt(3^2+1^2)}}}

{{{r=sqrt(10)}}}

{{{r=3.16}}}

add circle to the graph

{{{drawing(600, 600, -10, 10, -10, 10,
circle(0,-3,.12),circle(3/2,0,.12),circle(5,2,.12),circle(3,-2,.12),
locate(5,2,A(5,2)),locate(3,-2,B(3,-2)),circle(2,1,.12),locate(2,1,C(2,1)),
circle(2,1,3.16),
 graph(600, 600, -10, 10, -10, 10, 2x-3,-(1/2)x+2)) }}}


and, your equation of the circle is: 

{{{(x-h)^2+(y-k)^2=r^2}}}  where {{{h}}} and {{{k }}} coordinates o the center, {{{ r }}}is radius

plug in values:C({{{2}}},{{{1}}})=>{{{h=2}}} and {{{k=1 }}} and {{{r^2=10}}}


{{{(x-2)^2+(y-1)^2=10}}}