Question 980038
an equation of the line containing the centers of the two circles: 

{{{x^2 + y^2 - 10x - 10y + 49 = 0}}} ...........complete squares and write equation of a circle in a form {{{(x-h)^2+(y-k)^2=r^2}}}
{{{(x^2 - 10x+b^2) -b^2+ (y^2 - 10y+b^2) -b^2+ 49 = 0}}}

....recall, {{{(a-b)^2=a^2-2ab+b^2}}}
in your case {{{(x^2 - 10x+b^2)}}}, {{{a=1}}} and {{{2ab=10}}}, so we have {{{2*1*b=10}}}=>{{{2b=10}}}=>{{{b=5}}}

and we can write

{{{(x^2 - 10x+5^2) -5^2+ (y^2 - 10y+5^2) -5^2+ 49 = 0}}}

{{{(x - 5)^2 -25+ (y - 5)^2 -25+ 49 = 0}}}

{{{(x - 5)^2 + (y - 5)^2 -50+ 49 = 0}}}

{{{(x - 5)^2 + (y - 5)^2 -1 = 0}}}

{{{(x - 5)^2 + (y - 5)^2  = 1}}}

as you can see, {{{h=5}}},{{{k=5}}}, and {{{r=1}}}

so, the center is at ({{{5}}},{{{5}}})


now do same with other circle:

{{{x^2 + y^2 - 4x - 6y + 9 = 0}}}

{{{(x^2  - 4x+b^2)-b^2+ (y^2 - 6y +b^2)-b^2+ 9 = 0}}}......{{{2ab=4}}}=>{{{2*1*b=4}}}=>{{{b=2}}} and for {{{y}}} we have {{{2ab=6}}}=>{{{2*1*b=6}}}=>{{{b=3}}}

{{{(x^2  - 4x+2^2)-2^2+ (y^2 - 6y +3^2)-3^2+ 9 = 0}}}

{{{(x  - 2)^2-4+ (y - 3)^2-9+ 9 = 0}}}

{{{(x  - 2)^2+ (y - 3)^2-4 = 0}}}

{{{(x  - 2)^2+ (y - 3)^2= 4}}}

as you can see, {{{h=2}}},{{{k=3}}}, and {{{r=2}}}

so, the center is at ({{{2}}},{{{3}}})

now we have two points, ({{{5}}},{{{5}}}) and ({{{2}}},{{{3}}}), and we can find the equation of a line passing through these two points


*[invoke change_this_name10094 5, 5, 2, 3]

since your line is {{{y=0.666666666666667x + 1.66666666666667}}}

or {{{y=(2/3)x + 5/3}}}, we can multiply both sides by {{{3}}}

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

{{{3y=2x + 5}}}, in standard form will be

{{{2x-3y+5=0}}}

so, your answer is:

d. {{{2x - 3y + 5 = 0}}}


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