Question 1205414


{{{(x+2)^2 + (y-5)^2=4}}}.....eq.1

{{{-4x+y=k}}}......eq.2
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start with

{{{(x+2)^2 + (y-5)^2=4}}}.....eq.1, this is a circle with center at ({{{-2}}},{{{5}}}) and radius {{{r=2}}}
 
and find intercepts

the {{{y}}}-intercepts can be found by setting {{{x=0 }}}in the equation and solving for {{{y}]}:

{{{(0+2)^2 + (y-5)^2=4 }}}
{{{4 + (y-5)^2=4 }}}
{{{(y-5)^2=0}}}
{{{y-5=0}}}
{{{y=5}}}

=> {{{y}}}- intercept is at ({{{0}}},{{{5}}})

since {{{x }}}is squared, there must be one more point ({{{x}}},{{{5}}}), so

substitute {{{y=5}}} in {{{(x+2)^2 + (y-5)^2=4}}}.....eq.1 and solve for {{{x }}}to get one more point on the circle: 

{{{(x+2)^2 + (5-5)^2=4 }}}
{{{(x+2)^2 =4 }}}
{{{x+2=sqrt(4 )}}}
{{{x+2=2}}} or {{{x+2=-2}}}
{{{x=0}}} or {{{x=-4}}}

solutions:
{{{x=0}}}, {{{y=5 }}}=> point ({{{0}}},{{{5}}})
{{{x=-4}}}, {{{y=5}}}=> second point ({{{-4}}},{{{5}}})


since solutions should be same for both eq.1 and eq.2, then go to

{{{-4x+y=k}}}......eq.2, substitute {{{x }}}and {{{y}}} values from points ({{{-2}}},{{{5}}}) , ({{{0}}},{{{5}}})  and  ({{{-4}}},{{{5}}}) 

{{{-4*-2+5=k}}}
{{{k=13}}}


{{{-4*0+5=k}}}
{{{k=5}}}


{{{-4*-4+5=k}}}
{{{k=21}}}

so, maximum value of {{{k}}} is {{{21}}}