Question 1204888
the standard equation of the circle:

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


use point A ({{{7}}}, {{{2}}})

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

{{{h^2 - 14h + k^2 - 4k - r^2 = -53}}}...solve for {{{r}}}

{{{r = sqrt(h^2 - 14h + k^2 - 4k + 53)}}}....eq.1


use point B ({{{1}}}, {{{4}}})


{{{(1-h)^2+(4-k)^2=r^2}}}

{{{h^2 - 2h + k^2 - 8k - r^2 = -17}}}

{{{r = sqrt(h^2 - 2h + k^2 - 8k + 17)}}}....eq.2


 and, use point  C ({{{-7}}},{{{ 0}}})


{{{(-7-h)^2+(0-k)^2=r^2}}}

{{{h^2 + 14h + k^2 - r^2 = -49}}}

{{{r = sqrt(h^2 + 14h + k^2 + 49)}}}....eq.3


from eq.1 and eq.2 we have

{{{sqrt(h^2 - 14h + k^2 - 4k + 53)=sqrt(h^2 - 2h + k^2 - 8k + 17)}}}...square both sides

{{{h^2 - 14h + k^2 - 4k + 53=h^2 - 2h + k^2 - 8k + 17}}}

{{{h^2 - 14h + k^2 - 4k + 53-h^2 +2h - k^2 + 8k - 17=0}}}

{{{-12h + 4k + 36 = 0}}}...solve for {{{k}}}

{{{k = 3h - 9}}}......(1)


from eq.2 and eq.3 we have


{{{sqrt(h^2 - 2h + k^2 - 8k + 17)= sqrt(h^2 + 14h + k^2 + 49)}}}

{{{h^2 - 2h + k^2 - 8k + 17= h^2 + 14h + k^2 + 49}}}

{{{h^2 - 2h + k^2 - 8k + 17- h^2 - 14h - k^2 - 49=0}}}

{{{-16h - 8k - 32 = 0}}}

{{{k = -2h - 4}}}.....(2)



from (1) and (2) we have

{{{3h - 9=-2h - 4}}}

{{{3h +2h=9 - 4}}}

{{{5h=5}}}

{{{h=1}}}


go to

{{{k = -2h - 4}}}.....(2), substitute {{{h}}}

{{{k = -2 *1 - 4}}}

{{{k=-2-4}}}

{{{k=-6}}}

=> the center is at ({{{1}}},{{{-6}}})


go to


{{{r = sqrt(h^2 + 14h + k^2 + 49)}}}....eq.3, substitute {{{h }}}and{{{ k}}}

{{{r = sqrt(1^2 + 14*1 + (-6)^2 + 49)}}}

{{{r = sqrt(100)}}}

{{{r = 10}}}


equation of your circle is:

{{{x-1)^2+(y-(-6))^2=10^2}}}

{{{(x-1)^2+(y+6)^2=100}}}


{{{ drawing( 600, 600, -15, 15, -20, 15,
circle(7,2,.14), locate(7,2,A(7,2)),
circle(1,4,.14), locate(1,4,B(1,4)),
circle(-7,0,.14), locate(-7,0.98,C(-7,0)),
circle(1,-6,.14), locate(1,-6,O(1,-6)),
graph( 600, 600, -15, 15, -20, 15, -sqrt(-x^2+2x+99)-6, sqrt(-x^2+2x+99)-6)) }}}