Find the parametric equations for the line tangent to the circle at the point .

Question

Find the parametric equations for the line tangent to the circle  at the point  .

Quizplus · Accepted Answer

The circle centered at point $(-2,1)$ with radius 3 has equation: $$(x+2)^2 + (y-1)^2 = 9.$$

Clearly, the point of tangency occurs at $(-2, 4).$

To find the equation of the tangent line, we need to differentiate. The implicit equation of the circle can be treated as a function: $$y = f(x) = \sqrt{9 - (x+2)^2} + 1.$$

Applying the chain rule, we get: $$\frac{\text{d}y}{\text{d}x} = \frac{-2(x+2)}{2\sqrt{9-(x+2)^2}}.$$

At the point $(-2,4),$ we have $\frac{\text{d}y}{\text{d}x} = 0.$ Therefore, the tangent line is horizontal, which means the parameter $t$ is constant, say $t=t_0.$

Hence, we get a vector equation for the tangent line as follows: $$\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} -2 \ 4 \end{pmatrix} + t_0 \cdot \begin{pmatrix} 1 \ 0 \end{pmatrix}.$$

Therefore, the parametric equations of the tangent line are: $$\begin{aligned} x &= -2 + t_0 \ y &= 4. \end{aligned}$$

This is exactly the same form as in {\bf (C)}, so our answer is $\boxed{\textbf{(C)}}.$

Find the Parametric Equations for the Line Tangent to the Circle