Question 1015160
{{{(x+2k)^2 + (y-3k)^2 = 25}}} pass through the point ({{{1}}},{{{0}}})
find value of {{{k}}}:

{{{(x+2k)^2 + (y-3k)^2 = 25}}}....substitute {{{x}}} and {{{y}}} with coordinates of given point

{{{(1+2k)^2 + (0-3k)^2 = 25}}}....solve for {{{k}}}

{{{1+4k+4k^2 + 9k^2 = 25}}}

{{{13k^2+4k = 25-1}}}

{{{13k^2+4k -24=0}}}........use quadratic formula

{{{k = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{k = (-4 +- sqrt( 4^2-4*13*(-24) ))/(2*13) }}}

{{{k = (-4 +- sqrt( 16+1248 ))/26 }}}

{{{k = (-4 +- sqrt( 1264 ))/26 }}}

{{{k = (-4 +- sqrt( 16*79 ))/26 }}}

{{{k = (-4 +- 4sqrt( 79 ))/26 }}}......simplify

{{{k = (-2 +- 2sqrt( 79 ))/13 }}}

{{{k = (-2/13 +- 2sqrt( 79 )/13 )}}}

exact solutions:

{{{k = -2/13 + 2sqrt( 79 )/13 }}}

and

{{{k = -2/13 - 2sqrt( 79 )/13 }}}

approximate solutions:

{{{k = -2/13 + 2sqrt( 79 )/13 }}}=>{{{k = -0.1538461538461538 + 0.1538461538461538(8.9) }}}=>{{{k=1.21538461538461502}}}=>{{{k=1.22}}}

and

{{{k = -2/13 - 2sqrt( 79 )/13 }}}=>{{{k = -0.1538461538461538 -0.1538461538461538(8.9)  }}}=>{{{k= -1.523}}}


so, you have a circle 
{{{(x+2*1.22)^2 + (y-3*1.22)^2 = 25}}}
{{{(x+2.44)^2 + (y-3.66)^2 = 25}}}

and circle

{{{(x+2*(-1.523))^2 + (y-3*(-1.523))^2 = 25}}}
{{{(x-3.046)^2 + (y+4.569)^2 = 25}}}

and both should pass through the point ({{{1}}},{{{0}}})

check it:

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,0,.12),locate(1,-0.5,p(1,0)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(-(x+2.44)^2 +25)+3.66 , -sqrt(-(x+2.44)^2 +25)+3.66,sqrt(-(x-3.046)^2+25)-4.569,-sqrt(-(x-3.046)^2+25)-4.569)) }}}