A Handy Way to Find the Equation of a Tangent Line to a Circle

Posted: April 4, 2024 in Mathematics
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The DSE (click here to see what HKDSE is) is close. The math core exam is less than 2 weeks time. I am going to share a really nice equation to find the tangent of a given point on the circle. Suppose that the general form of a circle is given by

x^2 + y^2 + Dx + Ey + F = 0

then the equation of a tangent line at point (x_1, y_1) will be as follows:

\displaystyle x_1 x + y_1 y + D\left(\frac{x+x_1}{2}\right) + E\left(\frac{y+y_1}{2}\right) + F = 0

Proof:

Differentiate the equation of the circle with respect to x implicitly.

2x+2yy'+D+Ey'=0

hence y’ is,

\displaystyle y' = \frac{-2x-D}{2y+E}

then the slope of the tangent at point (x_1, y_1) would be

\displaystyle m=\frac{-2x_1-D}{2y_1+E}

By point slope form of a line,

\displaystyle y-y_1 = \frac{-2x_1-D}{2y_1+E}\cdot(x-x_1)

(y-y_1)(2y_1+E) = (x-x_1)(-2x_1-D)

2y_1y-2y_1^2+Ey-Ey_1 = -2x_1x+2x_1^2-Dx+Dx_1

2x_1x+2y_1y+D(x-x_1)+E(y-y_1) = 2x_1^2+2y_1^2

\displaystyle x_1x+y_1y+\frac{D(x-x_1)}{2}+\frac{E(y-y_1)}{2} = x_1^2+y_1^2

Now, since the given point satisfies the circle, we plug (x_1,y_1) into the circle,

x_1^2 + y_1^2 + Dx_1 + Ey_1 + F = 0

x_1^2 + y_1^2 = -Dx_1 - Ey_1 - F

Hence, we have the following,

\displaystyle x_1x+y_1y+\frac{D(x-x_1)}{2}+\frac{E(y-y_1)}{2} = -Dx_1 - Ey_1 - F

Move everything to the left, therefore

\displaystyle x_1 x + y_1 y + D\left(\frac{x+x_1}{2}\right) + E\left(\frac{y+y_1}{2}\right) + F = 0

QED. Good luck.

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