Question

How do you write an equation of the line tangent to \[{{x}^{2}}+{{y}^{2}}=169\] at the point \[(5,12)\]?

Answer 1

Hint: From the given question we are asked to find the equation of a line. We are given the equation of a circle, using this circle equation along with we will use the concept of implicit differentiation and find the slope of the tangent to the circle and at last we will use the equation of the line with slope and passing through point formula and solve the given question.

Complete step by step answer:
Here we are given the equation of the circle \[{{x}^{2}}+{{y}^{2}}=169\], we can also write this as follows.
\[\Rightarrow {{x}^{2}}+{{y}^{2}}=169\]
\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{13}^{2}}\]
To determine the slope of a tangent to the circle at any point we need to use the implicit differentiation.
\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{13}^{2}}\]
\[\Rightarrow 2x+2y\dfrac{dy}{dx}=0\]
Here we used the formula in differentiation which is \[\Rightarrow \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\].
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}\]
At the point \[(5,12)\], it will be as follows.
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{5}{12}\]
So the tangent has a slope of \[-\dfrac{5}{12}\] and passes through the point \[(5,12)\].
We know that the general form of an equation of line with slope m and passing through \[(x,{{x}^{'}})\] a point will be as follows.
\[\Rightarrow (y-x)=m(x-{{x}^{'}})\]
So, now we will substitute our values in the general form and simplify the equation.
So, we get the equation reduced as follows.
\[\Rightarrow (y-12)=-\dfrac{5}{12}(x-5)\]
\[\Rightarrow (y-12)12=-5(x-5)\]
\[\Rightarrow (12y-144)=-(5x-25)\]
\[\Rightarrow 12y=-5x+25+144\]
\[\Rightarrow y=\dfrac{-5x+169}{12}\]

Note: Students must be very careful in doing the calculations. Students should have good knowledge in the concept of differentiation and general form of line.
We must know the formulae given below to solve our question.
\[\Rightarrow (y-x)=m(x-{{x}^{'}})\](general form of line with slope and point given)
\[\Rightarrow \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\] (basic formula in differentiation)