Question

How do you find the equation of the tangent to the curve ${{x}^{2}}+{{y}^{2}}=169$ at the point $(5,-12)$ ?

Answer 1

In the above question we are asked to determine the tangent to the given curve. As we can see that the given curve is a circle passing through origin and having radius equal to 13. In order to find the equation of the given the tangent first we need to find the slope of the tangent. Then by using slope intercept form of equation we will get the equation of the tangent at the point $(5,-12)$.

Complete step by step answer:
In the above question the equation of the given circle is ${{x}^{2}}+{{y}^{2}}=169$.Here we can see that the centre of the circle is origin and its radius is equal to 13. Now in order to determine the slope of the tangent to the circle at any point we will find the value of the gradient of the normal at the point of contact.
The formula to find the gradient of the normal is $m=\dfrac{x-{{x}_{1}}}{y-{{y}_{1}}}$
$\Rightarrow m=\dfrac{12-0}{5-0}=\dfrac{12}{5}$
As we know that the radius is perpendicular to the tangent therefore, it act as a normal to the tangent.
We know that for perpendicular lines the product of their gradients is -1.
Using the above formula, we will find the slope of the tangent.
We know that

& {{m}_{1}}\times m=-1 \\\ & \Rightarrow {{m}_{1}}\times \dfrac{12}{5}=-1 \\\ & \Rightarrow {{m}_{1}}=-\dfrac{5}{12} \\\ \end{aligned}$$ Now we will use the slope intercept form of the equation to find the equation of the tangent. $$\begin{aligned} & y-{{y}_{1}}=m(x-{{x}_{1}}) \\\ & \Rightarrow y-12=\dfrac{-5}{12}(x-5) \\\ & \Rightarrow 12y-144=-5x+25 \\\ & \Rightarrow 12y+5x=169 \\\ \end{aligned}$$ Therefore, the equation of the tangent to the curve $${{x}^{2}}+{{y}^{2}}=169$$ at the point $$(5,-12)$$ is $$12y+5x=169$$ .  **Note:** In the above question instead of finding the gradient we can use implicit differentiation to determine the slope of the tangent. While solving the above question keep in mind the formulas used such as the product of slope and the point slope form of general equation. Perform the differentiation carefully in case of implicit differentiation method.