Question 1156565
The first thing I would do is draw a diagram and label some points.


<a href="https://www.imageupload.net/image/xBjtx"><img src="https://imagehost.imageupload.net/2020/04/20/lindsay1.1.th.gif" alt="lindsay1.1.gif" border="0" /></a>


Since the distance from the center of the circle {{{C}}} to {{{P}}} is {{{r}}}


    {{{(r -1)^2+ (3 -0)^2 = r^2}}} 

Solve for {{{r}}}.

{{{r^2-2r+1 + 9 = r^2}}}

{{{r^2-r^2 - 2r+1 + 9 = 0}}}

 {{{-2r+10 = 0}}}

{{{ 10 =2r}}}

{{{r=5}}}


yor equation is:

 {{{(x - h)^2 + (y-k)^2 = 5^2}}}.........use given point ({{{1}}},{{{0}}}) to find {{{h}}} and {{{k}}}

{{{(1 - h)^2 + (0-k)^2 = 25}}}

{{{1 - 2h+h^2 + k^2 = 25}}}

{{{ - 2h+h^2 + k^2 = 24}}}......eq.1


 {{{(x - h)^2 + (y-k)^2 = 5^2}}}.........use given point ({{{0}}},{{{3}}}) 

{{{(0 - h)^2 + (3-k)^2 = 25}}}

{{{h^2 + k^2-6k +9= 25}}}

{{{h^2 + k^2-6k = 16}}}..............eq.2

solve the system:

{{{ - 2h+h^2 + k^2 = 24}}}......eq.1
{{{h^2 + k^2-6k = 16}}}..............eq.2
---------------------------------------------------------subtract

{{{ - 2h+h^2 + k^2-(h^2 + k^2-6k) = 24-16}}}

{{{ - 2h+h^2 + k^2-h^2 - k^2+6k = 8}}}

{{{ - 2h +6k = 8}}}

{{{ - h +3k = 4}}}......solve for {{{h}}}

{{{ - 4+3k = h}}}.........eq.1a


go to

{{{h^2 + k^2-6k = 16}}}..............eq.2, substitute {{{h}}}

{{{(- 4+3k)^2 + k^2-6k = 16}}}......solve for {{{k}}}

{{{10 k^2 - 30 k + 16 = 16}}}

{{{10 k^2 - 30k  = 16-16}}}

{{{10 k^2 - 30k  = 0}}}

{{{ k^2 - 3k  = 0}}}

{{{k( k - 3)  = 0}}}


=>{{{k = 0}}} or{{{ k = 3}}}



go to {{{ - 4+3k = h}}}.........eq.1a, substitute {{{k}}}


{{{ - 4+3*0 = h}}}=>{{{h=-4}}}

{{{ - 4+3*3 = h}}}=>{{{h=5}}}



solutions for coordinates of the center
{{{h = -4}}}, {{{k = 0}}}=> disregard, because point ({{{0}}},{{{3}}}) is a point where circle touches y-axis

your coordinates of the center:

{{{h = 5}}}, {{{k = 3}}}


and your equation is:


{{{(x - 5)^2 + (y-3)^2 = 25}}}

{{{ graph( 600, 600, -10, 10, -10, 10,sqrt(-(x - 5)^2 +25) +3,-sqrt(-(x - 5)^2 +25)+3) }}}