I'm asking about a question about two lines which are tangents to a circle. Most of the question is quite elementary algebra, it's just one stage I can't get my head round. Picture here:

enter image description here

The circle $C$ has equation $(x-6)^2+(y-5)^2=17$. The lines $l_1$ and $l_2$ are each a tangent to the circle and intersect at the point $(0,12)$. Find the equations of $l_1$ and $l_2$ giving your answers in the form $y=mx+c$.

Both lines have equation $y = mx + 12$ where $m$ represents two gradients to be found (both negative).

The circle has equation $(x-6)^2 + (y-5)^2 = 17$.

Combining the knowledge $y = mx + 12$ for both lines and $(x-6)^2 + (y-5)^2 = 17$ produces the quadratic:

$$(1+m^2)x^2 + (14m−12)x + 68 = 0$$

At this point I was confused about what step to take to get $m$ or $x$. Looking at the worked solution, it says "There is one solution so using the discriminant $b^2 − 4ac = 0$..."

From here it's straightforward algebra again, producing another quadratic based on the $b^2 - 4ac$ of the previous quadratic:

$$(14m−12)^2 - 4 x (1+m^2) + 68 = 0$$

etc. until we have $m = -4$ or $-8/19$.

My question is I don't understand how we can tell it's right to assume $b^2 - 4ac = 0$ and how we can see that's the right step to take in this question.

Obviously this feels intuitively wrong since we know there are two solutions for m. Is the logic that m is a gradient which intersects with the circle once? But if so, how do you see that this is the right equation to decide it only has one solution? (I had assumed before looking at the worked example if I needed to do something more complicated based on the equations of the radii or the knowledge that the two tangents would be equal length from the circle to where they meet or something.)

As you can tell, this question is based on fairly elementary algebra; I'm more concerned about knowing why this is the right step to take here.

Many thanks for any answers.

  • 6
    $\begingroup$ Each tangent line intersects the circle at exactly one point, hence discriminant equals zero $\endgroup$ Aug 15, 2021 at 9:40
  • $\begingroup$ Please use MathJax in to format math expressions properly. $\endgroup$
    – g.kov
    Aug 15, 2021 at 10:18

6 Answers 6


When you insert $y=mx+12$ into the circle equation and obtain $$(1+m^2)\cdot x^2 + (14m−12)\cdot x + 68 = 0\label{1}\tag{1}$$ There are two things you can do: solve for $x$ or for $m$. Which of those makes sense?

Notice that in $(\ref{1})$, if you take $x$ as a given constant and solve for $m$, you are answering the question: Given an $x$ intercept of the line through $(0,12)$ with the circle, determine the slope of this line. It might happen that there are two points, one or none on the circle with that given $x$ coordinate, hence the number of solutions for $m$. This isn't, however, what we are looking for.

Instead, the tangent-condition offers us valuable information: there is only one $x$ solution for some $m$ we are looking for. Thus, it makes more sense to solve for $x$ in $(\ref{1})$ with the quadratic equation formula: $$x = \frac{- 7 m + 6\pm\sqrt{-19 m^2 - 84 m - 32} }{m^2 + 1}$$ This might look frightening at a first glance, but remember that we already know that there is exactly one solution for $x$, and this happens precisely when the discriminant is zero: $$\Delta=-19m^2-84m-32=0$$ Which can be solved via the quadratic equation formula. This yields $$\fbox{$\displaystyle m\in\left\{-4, -\frac{8}{19}\right\}$}$$


Just to restate the essential point so that it doesn't get buried:

  • The solutions to the equation correspond to intersection points between the line and the circle. There can be zero, one or two such points.
  • If the line does not touch the circle, there are no (real) solutions.
  • If the line passes through the interior of the circle, there are two solutions.
  • Only if the line is tangent to the circle, i.e. only touches it at one point, will the equation have only one solution. Since we want to find lines that are tangent to the circle, this is exactly the case that we're interested in.

Conveniently, we also (should) know that a quadratic equation has only one solution if and only if its discriminant is zero. So we don't actually need to solve the original equation — it's enough to find the value(s) of $m$ for which the discriminant is zero. But that's just an algebraic shortcut. If we didn't know this fact already, we could've obtained the same result slightly more laboriously by first applying the quadratic formula to find the (up to) two solutions, and then finding the value(s) of $m$ for which they are equal (i.e. actually just one solution).

  • 4
    $\begingroup$ I feel like this is the only answer that really addresses the issue being asked by OP. The other ones focus on the rest of the resolution rather than the why. $\endgroup$ Aug 16, 2021 at 17:28
  • $\begingroup$ If we consider Y axis As passing through the centre of circle and solve it with the circle,Still The discriminant is 0,How can we always say that a line is tangent if Discriminant=0 ,I mean for X²+Y²=4,and line x=0,If we find discriminant ,Still it is 0?? $\endgroup$ May 28 at 22:02
  • $\begingroup$ @DheerajGujrathi: Substituting $x = 0$ into $x^2 + y^2 = 4$ gives you $y^2 = 4$, or in standard form, $y^2 - 4 = 0$. What's the discriminant for that quadratic? (Hint: It's not zero.) $\endgroup$ May 29 at 16:26

The intersection point(s) of two curves is given by the solution(s) of their simultaneous equations.

In particular, when a straight line $$y=mx+c\tag1$$ and a circle $$(x−a)^2+(y−b)^2=r^2\tag2$$ intersect at exactly one point, i.e., are tangent to each other, their combined equation$$(x−a)^2+(mx+c−b)^2=r^2\\(m^2+1)x^2+2(mc-mb-a)x+(a^2+b^2+c^2-r^2-2bc)=0$$ has exactly one solution. But since this equation is a quadratic, its discriminant $$4(mc-mb-a)^2-4(m^2+1)(a^2+b^2+c^2-r^2-2bc)\tag3$$ equals $0.$

For our example, plug $c=12,a=6,b=5, r^2=17$ into expression $(3)$ and set it to $0:$ $$4(7m-6)^2-4(m^2+1)(68)=0\\19m^2+84m+32=0\\m=-4 \;\:\text{ or } -\frac8{19}.$$


An approach using geometry:

Let $P = (0,12)$ and the two tangent points be $A, B$. Then $OP^2 = (0-6)^2 + (12-5)^2 = 85$, and the radius $AO^2 = 17$, so $PA^2 = PB^2 = 68$ as tangents to the circle are the same length.

Draw a line connecting $A$ and $B$, and call its intersection with $OP$, $M$. By similarity ($\Delta PAO \sim \Delta AOM$) and using power of a point, $(2k)^2 = (\sqrt{17} + k)(\sqrt{17} - k) \implies 4k^2 = 17 -k^2 \implies k^2 = 17/5$, where $k = AM = BM$.

The line $OP$ has gradient $\frac{5-12}{6-0} = -\frac{7}{6}$, so the equation of the line is $y = -\frac{7}{6}x + 12$. Making a right triangle with $OM$ as the hypotenuse, $\text{run}^2 + \text{rise}^2 = 17/5$, or $x^2 + \left(\frac{7}{6}x\right)^2 = 17/5 \implies x = -\frac{6}{5}$, so $M$ has $x$-coordinate $6 - \frac{6}{5} = \frac{24}{5}$ and $y$-coordinate $\frac{32}{5}$.

Also, since $OP \perp AB$ using another circle theorem, $AB$ has gradient $\frac{6}{7}$ and $AM$ has length $2k$. Drawing another right triangle with hypotenuse $AM$, $x^2 + \left(\frac{6}{7}x\right)^2 = 68/5 \implies x = -\frac{14}{5}$, so $A$ has coordinates $\left(\frac{24}{5} - \frac{14}{5}, \frac{32}{5} - \frac{14}{5} \cdot \frac{6}{7} \right) = (2,4)$. Now $BM$ has the same length in the opposite direction, so $B = \left(\frac{24}{5} + \frac{14}{5}, \frac{32}{5} + \frac{14}{5} \cdot \frac{6}{7} \right) = \left(\frac{38}{5}, \frac{44}{5} \right)$.

Now use point-slope form to find the equation of the lines $PA,PB$, and you will get $-\frac{8}{19}x + 12, -4x + 12$.

Desmos graph is here.

  • $\begingroup$ Id have used 2 radii, not a chord, and solved for the angle at the intercept. Then the tangents are essentially at +/- that. It keeps the maths simpler. (Also any but basic circle geometry theorems à la Euclid, arent syllabus knowledge so much these days) $\endgroup$
    – Stilez
    Aug 17, 2021 at 2:02
  • $\begingroup$ Of course you can do this and there are many other ways, but you would need trigonometry and the tan angle addition formulas. My solution does not use trigonometry. $\endgroup$
    – Toby Mak
    Aug 17, 2021 at 3:33
  • 1
    $\begingroup$ Minimal trigonometry (right angle triangles only), and no need for tan addition formula; just being aware that gradient is tan of (angle to horizontal), which is fairly well taught. No complaint, multiple approaches let the OP choose whichever is best for them, and see there are diverse ways to solve most problems. Posted my approach below if its of interest. $\endgroup$
    – Stilez
    Aug 17, 2021 at 12:12

Other people have explained in depth your question about the determinant, and why it's zero. In summary:

When you substitute the line's equation into the circle's equation, the resulting equation says in effect "given a line with formula y=12-mx, for some arbitrary value "m", what values of x does the line intersect/touch the circle?"

The values of "m that we want are those making the line a tangent. But what defines a tangent?

For any given value of "m", that is a tangent to the circle, the line must touch the circle at exactly one point. Not zero points, not 2 points. Just one.

That means the quadratic needs to have a zero determinant, otherwise it would be saying the line crosses the circle at no points (negative determinant, no real root), or 2 points (positive determinant, 2 real roots). So you want values of "m" that make the determinant zero.

You then get an equation for "m", which gives you "gradients where the line will be a tangent to the circle". Its a quadratic because two values of m can do this.

Geometrically, there is a very intuitive solution however, using symmetry

Draw the diagram like this

Edited pic

(Call the distance from the intercept to the centre of the circle, D, I forgot to mark that)

We can immediately see that

  • the two tangents are symmetrical in the sense shown.
  • from the equation, the circle's centre and radius are (6,5) and √17.
  • the distance from the intercept (0,12) to the circle centre (6,5) is D = √(62+(12-5)2) = √85

Now let's find the angle ɑ. We have a right angle triangle so ɑ = sin-1(D/r) = sin-1(√17/√85) = sin-1(1/√5) = 26.565 degrees.

Last, let's find the angle (not gradient!) of the line between the intercept and the center of the circle. The gradient is (5-12)/6 = -7/6 so the angle is tan-1(-7/6) = -49.399 degrees.

By construction, the two triangles are identical sides and angles, so the two gradients are lines that pass through (0,12) at angles -49.399 ± 26.565 degrees, or -75.965 and -22.834 degrees.

Converting the angles of the lines back to gradients, we get that the tangents have gradients: tan(-75.965) and tan(-22.834), or -4 and -0.421 (which as you expected, is -8/19).

We know that both tangents have lines with formulae y = 12 - mx, because they are straight lines passing through (0,12). Note that ive used "-m" for clarity here, we expect two negative numbers, although mathematically its cleaner to use "+mx".

So the tangents have equations:

y = 12 - 4 x


y = 12 - 0.421 x

Not a quadratic in sight!

In general, converting questions about tricky tangents to circles, into questions about right angle triangles based on a radius and a tangent to a circle, is a trick that should be in any mathematician's toolbox.


after combining/making substitutions for the equations of a generic line and the circle in the given, 'there's one solution in x' means that the line and the circle only have 1 point of intersection. and that's actually part of the given: 'tangent lines'

graphically, one can confirm that only 2 such lines exist. and when solving for m, this fact should be reflected as well


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