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Given the endpoints $(11, 23)$ and $(6, 13)$ of a circle, find the equation of the circle and the equation of a line tangent to the circle.

First, I found the center using the midpoint formula:

$$ \left(\frac{11+6}{2}, \frac{13+23}{2}\right)$$ $$\left(\frac{17}{2}, 18\right)$$

Then I found the radius by finding the distance between the points and dividing by 2:

$$\frac{\sqrt{(13 - 23)^2 + (6 - 11)^2}}{2} = \frac{5\sqrt{5}}{2}$$

The equation:

$$(x - 17/2)^2 + (y - 18)^2 = \left(\frac{5\sqrt{5}}{2}\right)^2$$

How can I find an equation tangent to this? I tried finding the slope between $(11, 23)$ and $(6, 13)$, finding the negative reciprocal, and plugging in one of the points into the point slope form, but when I graphed it, the line wasn't tangent to the circle.

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    $\begingroup$ I'm a bit confused as to why it asks for the tangent line. Because there's not just one. $\endgroup$ – Ataraxia Apr 26 '13 at 0:34
  • $\begingroup$ Right... I edited it. $\endgroup$ – badjr Apr 26 '13 at 0:35
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Perhaps you should find the tangent line at the two given points $(11, 23)$ and $(6, 13)$: Since they are endpoints of a diameter of the circle, the slopes of the tangent lines to the circle at each point, respectively, will be equal (they will be parallel).

The slope of the line connecting them will thus be perpendicular to the tangent lines at those points.

So find the slope of the line connecting the two given points. Call it $m$. We see $\quad m = \dfrac{23-10}{11 - 5} = 2$

The slope of the lines tangent to the circle at the given points will then be $-\dfrac 1m = -\dfrac 12$.

Then you can find both the equation of the line tangent to the circle at $(11, 23)$, and the equation of the line tangent to the circle at $(6, 13)$, using the point-slope form for the equation of a line. E.g., the tangent to the circle at $(6, 13)$ will be $$y - 13 = -\dfrac 12(x - 6) \iff y = -\dfrac 12 x + 16\qquad\qquad\tag{tangent line at $(6, 13)$}$$

Indeed, the process outlined above sounds like what you tried, so I suspect you made an error somewhere in your calculation of slope, or forming the equation of one of the tangent lines:

Graphing both the circle and the equation of the line given above in Wolfram Alpha gives this nice tangent to the circle

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  • $\begingroup$ Nice descriptions and a slow night indeed! + 1 $\endgroup$ – Amzoti Apr 26 '13 at 1:32
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The simplest tangent lines would be horizontal or vertical. Try looking for those.

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As vadim123 said, finding the horizontal and vertical tangent lines is trivial. To find any tangent line on the circle, take the derivative of the parametric equations: and divide them to find the slope of a tangent line: $$y-y_1=m(x-x_1)$$ $$\frac{d}{d\theta}x=-sin(\theta)$$ $$\frac{d}{d\theta}y=cos(\theta)$$ $$m = -\frac{cos(\theta)}{sin(\theta)}$$

here $\theta$ can be any angle. To find $x_1$ and $y_1$, use the same angle $\theta$ and plug it into $x=\frac{5\sqrt{5}}{2}cos(\theta)+17/2$ and $y=\frac{5\sqrt{5}}{2}sin(\theta)+17/2$.

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  • $\begingroup$ As you can see, $tan$ is called the tangent function for a reason. $\endgroup$ – Ataraxia Apr 26 '13 at 1:23
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Using vadim's idea: You know that the center is at $(17/2, 18)$, and the radius is $5\sqrt5/2$. So, the point $(17/2, 18 + 5\sqrt5/2)$ is at the "top" of the circle. The tangent at this point is horizontal; it is the line $y = 18 + 5\sqrt5/2$.

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The chords from the endpoints of a diameter of a circle to any other point on the circle form a right angle. Two nonzero vectors are at right angles to each other iff their dot product vanishes. These two facts allow one to immediately write an equation for a circle given the endpoints $(x_1,y_1)$ and $(x_2,y_2)$ of a diameter: $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.\tag{*}$$ If you need to put this in the more familiar form $(x-x_c)^2+(y-y_c)^2=r^2$, it’s a simple enough matter to find the midpoint of the diameter, as you have done, and then rearrange (*) accordingly.

You can also find an equation of tangent line to this circle almost mechanically: if $P=(x_p,y_p)$ is a point on the circle with equation $(x-x_c)^2+(y-y_c)^2=r^2$, then an equation for the tangent at $P$ is $(x_p-x_c)(x-x_c)+(y_p-y_c)(y-y_c)=r^2$. If your equation is in the more general form $x^2+y^2+ax+by+c=0$, there’s a slightly more complicated substitution that you can make to convert it into an equation for the tangent at $P$: $$x^2\to xx_p, y^2\to yy_p, x\to\frac12(x+x_p), y\to\frac12(y+y_p).$$ (This substitution, with one more rule for the $xy$ term, works for the general equation of any conic, in fact.)

Either way, since you’ve already found an equation for your circle and know two points on it, finding an equation for the tangent at either of those points is a straightforward matter of substitution and simplification.


If you’re comfortable working with matrices, you can also use the fact that the polar of a point on a circle is the tangent at that point. Without going into details, this means that an equation of the tangent at $P$ is $$\begin{bmatrix}x&y&1\end{bmatrix} \begin{bmatrix}1&0&-x_c\\0&1&-y_c\\-x_c&-y_c&x_c^2+y_c^2-r^2\end{bmatrix} \begin{bmatrix}x_p\\y_p\\1\end{bmatrix}=0,$$ but this amounts to making the substitution into the circle’s equation that I described above. (This works for any conic, in fact.)

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