Question 1088210
The slope of the tangent line is equal to the value of the derivative at the point.
Complete the square to get to center-radius form of a circle,
{{{(x^2-22x)+(y^2+12y)=-137}}}
{{{(x^2-22x+121)+(y^2+12y+36)=-137+121+36}}}
{{{(x-11)^2+(y+6)^2=20}}}
So implicitly differentiating,
{{{2(x-11)dx+2(y+6)dy=0}}}
{{{(y+6)dy=(11-x)dx}}}
{{{dy/dx=(11-x)/(6+y)}}}
So at {{{x=15}}}, find the y value,
{{{15^2+y^2-22(15)+12y+137=0}}}
{{{225+y^2-330+12y+137=0}}}
{{{y^2+12y+32=0}}}
{{{(y+4)(y+8)=0}}}
So then,
{{{x=15}}}
{{{y=-4}}}
{{{m[1]=(11-15)/(6-4)=-4/2=-2}}}
and
{{{x=15}}}
{{{y=-8}}}
{{{m[2]=(11-15)/(6-8)=-4/-2=2}}}
So then using the point-slope form of the line,
{{{ y+4=-2(x-15) }}}
and
{{{ y+8=2(x-15) }}}
I leave it to you to convert to slope-intercept form.
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*[illustration pl5.JPG].