Question 946148
{{{x^2+y^2-8x-6y+9=0}}}
{{{(x^2-8x)+(y^2-6y)=-9}}}
{{{(x^2-8x+16)+(y^2-6y+9)=-9+16+9}}}
{{{(x-4)^2+(y-3)^2=16}}}
{{{(y-3)^2=16-(x-4)^2}}}
{{{y-3= 0 +- sqrt(16-(x-4))^2}}}
{{{y=-3 +- sqrt(16-(x-4))^2}}}
Now you can look at the values of {{{Z=y/x}}}
{{{Z=(-3 +- sqrt(16-(x-4)^2))/x}}}
Since Z is now the function of only one variable, take the derivative and set it equal to zero to find the maximum value.
We'll do it in two parts because of the plus/minus value.
{{{Z[1]=(-3 + sqrt(16-(x-4)^2))/x}}}
{{{Z[1]}}} has a minimum of {{{y/x=3/8}}} when {{{x=8}}}.
{{{Z[2]=(-3 - sqrt(16-(x-4)^2))/x}}}
{{{Z[2]}}} has a minimum of {{{y/x=-7/24}}} when {{{x=72/25}}}.
There is no maximum so the range of {{{y/x}}} is [{{{-7/24}}},{{{infinity}}})