Coming back to this after about 6 months I now know how to solve it.

First I found the gradient of the radius $\frac{changeiny}{changeinx}$ >> $\frac{-8}{6}$ >> $1.33333333$

Then I found the negative reciprocal of the radius gradient to get the gradient of the tangent because it is at $90^\circ$ to the radius which becomes $\frac{1}{1.33333333}$ >> $0.75$

I then substituted the $x$ and $y$ values where the tangent touches the circle into the $y=mx+c$ equation >> $-8=0.75*6+c$ >> $-8=4.5+c$ >> $c=-12.5$

Which means that the equation of the tangent turned out to be ..... $y=0.75x-12.5$


The circle has the equation $x^2+y^2=100$

Find the equation of the tangent to the circle at the point $A(6,-8)$.

Any clues on how to solve this?



closed as off-topic by Jack D'Aurizio, C. Falcon, Trevor Gunn, JonMark Perry, mlc Jun 7 '17 at 4:47

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  • $\begingroup$ What do you suppose the slope of that line is? $\endgroup$ – Doug M Jun 6 '17 at 22:53
  • $\begingroup$ Are you aware of implicit differentiation? Or you could just use nice points on the graph (e.g. integer values for point entries) to find the slope, and you have a point on the line (to find the intercept). $\endgroup$ – Dave Jun 6 '17 at 22:53
  • $\begingroup$ I'm still in high school. This was thrown in with a few circle theorem questions. I don't know what implicit differentiation is, sorry. $\endgroup$ – L Smith Jun 6 '17 at 22:54
  • $\begingroup$ I tried putting it into the format y = mx + c but this didn't follow the line when I tested it. $\endgroup$ – L Smith Jun 6 '17 at 22:55
  • $\begingroup$ Apply some similarity and find where such tangent meets the axis. $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 23:05

Since the shape is a circle, the tangent line at a point on the circle is perpendicular to the line connecting that point to the origin. The slope of the line from the origin to $A$ is clearly $\frac{-4}{3}$ since the "rise" from $(0,0)$ to $A=(6,-8)$ is $-8$ and the "run" is $6$. Thus, the slope of the perpendicular line is $\frac{3}{4}$. So the slope of your tangent line is $\frac{3}{4}$, and a point on the line is $A=(6,-8)$.


enter image description here

By straightforward similarities, the wanted line goes through $\left(6+\frac{32}{3},0\right)$ and $\left(0,-\frac{25}{2}\right)$.

There aren't many lines that do that.


hint: Use implicit differentiation to find $y' = -\dfrac{x}{y}|(6,-8) = \dfrac{3}{4}$. Can you take it from here? If you are not familiar with implicit differentiation, then use the line $OA$ whose equation is $y =-\dfrac{4}{3}x$. Thus the line perpendicular to it has slope $ \dfrac{3}{4}$. Can you take it from here as well?

  • $\begingroup$ I was hoping there was just something I was missing but it seems I'm a bit out of my level of knowledge here. $\endgroup$ – L Smith Jun 6 '17 at 22:58

Alternatively if you haven't learnt calculus you can say there a line in the form $y=mx+c$ passing through the point $(6,-8)$ so $-8= 6m+c \implies c=-8-6m \implies y=mx+(-8-6m) $

Then substitute this into the circle equation:

$$ x^2+y^2=100 \implies x^2+(mx+(8-6m))^2=100 $$

$$ \Leftrightarrow x^2+m^2x^2+2mx(8-6m)+(8-6m)^2=100 \implies m^2 x^2 - 12 m^2 x + 36 m^2 + 16 m x - 96 m + x^2 + 64=100$$

$$ \Leftrightarrow x^2(m^2+1) +x(-12m^2+16m)+(36m^2-96m-36)=0$$

For tangency the discriminant of this equation must equal zero so

$$ (-12m^2+16m)^2-4(m^2+1)(36m^2 - 96m-36) =0$$

$$ \Leftrightarrow 16(4m+3)^2=0 \implies m=\frac{-3}{4}$$

Hence the equation of the tangent line is

$$ y=\frac{-3}{4}x - \frac{7}{2}$$


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