Question 243003
First derive both sides of *[Tex \LARGE x^2+y^2=169] with respect to 'x' to get


*[Tex \LARGE x^2+y^2=169]


*[Tex \LARGE \frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(169\right)]


*[Tex \LARGE 2x+2yy^{\prime}=0]


Now solve for y':


*[Tex \LARGE 2x+2yy^{\prime}=0]


*[Tex \LARGE 2yy^{\prime}=-2x]


*[Tex \LARGE y^{\prime}=\frac{-2x}{2y}]


*[Tex \LARGE y^{\prime}=-\frac{x}{y}]



So the slope is simply the negative quotient of the two coordinates. For the point (5,-12), the slope at that point is {{{m=-5/(-12)=5/12}}} (since {{{x=5}}} and {{{y=-12}}})



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Now recall that the point slope formula is 


{{{y-y[1]=m(x-x[1])}}} where 'm' is the slope and *[Tex \LARGE \left(x_{1},y_{1}\right)] is the point in which the line goes through.



{{{y--12=(5/12)(x-5)}}} Plug in {{{m=5/12}}}, {{{x[1]=5}}}, and {{{y[1]=-12}}}



{{{y+12=(5/12)(x-5)}}} Rewrite {{{y--12}}} as {{{y+12}}}



{{{y+12=(5/12)x+(5/12)(-5)}}} Distribute {{{5/12}}}



{{{y+12=(5/12)x-25/12}}} Multiply {{{5/12}}} and {{{-5}}} to get {{{-25/12}}}



{{{y=(5/12)x-25/12-12}}} Subtract 12 from  both sides to isolate y



{{{y=(5/12)x-169/12}}} Combine like terms {{{-25/12}}} and {{{-12}}} to get {{{-169/12}}} 



{{{12y=5x-169}}} Multiply EVERY term by the LCD 12 to clear out the fractions.



{{{-5x+12y=-169}}} Subtract 5x from both sides.



{{{5x-12y=169}}} Multiply EVERY term by -1 to make the 'x' coefficient positive.

 

So the equation of the tangent line is {{{5x-12y=169}}}