📂Geometry

Finding the Equation of the Tangent Line at a Point on a Circle

Explanation

Let’s find the equation of the tangent line at a point $(x_{1},y_{1})$ on the circle $x^2+y^2=r^2$. This can be divided into cases when $y_{1}\neq 0$ and when $y_{1}=0$.

$y_{1}\neq 0$

The slope from the center of the circle to the tangent point is $\dfrac{y_{1}}{x_{1}}$. Since the product of the slopes of two perpendicular lines is -1, the slope of the tangent line is $-\dfrac{x_{1}}{y_{1}}$. The equation of the line passing through point $(x_{1},y_{1})$ with a slope $-\dfrac{x_{1}}{y_{1}}$ is

$$y-y_{1}=-\frac{x_{1}}{y_{1}}(x-x_{1})$$

$$\implies y_{1}y-y_{1}^2=-x_{1}x+x_{1}^2$$

$$\implies x_{1}x+y_{1}y=x_{1}^2+y_{1}^2=r^2$$

Therefore, the equation of the tangent line when $y_{1}\neq 0$ is

$$x_{1}x+y_{1}y=r^2$$

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$y_{1}=0$

As you can see in the figure, when $(x_{1},0)$, $x=x_{1}=\pm r$ applies. However, when substituting $y_{1}=0$ into the equation of the tangent line of $y_{1}\neq 0$, the same form appears. That is, the same equation applies whether it is $y_{1}\neq 0$ or $y_{1}=0$. Therefore, the equation of the tangent line at a point $(x_{1},y_{1})$ on the circle $x^2+y^2=r^2$ is $x_{1}x+y_{1}y=r^2$.