How do I find the tangent line for these two points?

Question

I need to find the slope of the tangent line to the circle, but I forgot how to find the tangent line for each of the points (0.8, -0.6) and (0.8, 0.6)

Answer 1

This can be easily solved using implicit differentiation.

The equation of the unit circle is $x^2+y^2=1$; take the derivative of both sides with respect to $x$, we get $\frac{\text{d}}{\text{d}x}(x^2+y^2)=\frac{\text{d}}{\text{d}x}1$, which gives $2x+2y\frac{\text{d}y}{\text{d}x}=0$ from chain rule. Rearranging gives $\frac{\text{d}y}{\text{d}x}=-\frac x y$. Substitute in those coordinates, we get $m_1=-\frac{4}{3}, m_2=\frac{4}{3}$.

However, since you posted this question in the geometry tag, I will provide an alternative approach, using only coordinate geometry.

The equation of the line passing through the origin $O(0,0)$ and a point on the circle $A(x,y)$ has a slope $\frac{y}{x}$. From the property of circles, we know that the tangent is perpendicular to the radius, which means the product of their slopes is $-1$. Suppose the slope of the tangent is $m$, we get $m\frac{y}{x}=-1\implies m=-\frac x y$. The result is the same as in the calculus method.

Answer 2

You can find the slope directly with the derivative like Joshua suggests.

Another way is to note that the tangent is perpendicular to the radius:

• Find the center of the circle.
• Calculate the slope of the radius to each point.
• From these, calculate the slope of the perpendicular.

Can you follow this and take it home from here?