Question 1089604
{{{x^2 +y^2 -4x-6y-3 =0}}} is ({{{h}}},{{{k}}}) and the radius is {{{r}}}, then {{{h+k+r}}}=?

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

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

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

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

{{{(x -2)^2 +(y-3)^2=16}}}

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

=> {{{h=2}}}, {{{k=3}}}, {{{r=4}}}

then {{{h+k+r=2+3+4=9}}}



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