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Is there an easy geometric way to prove that the circle is tangent to the line $\overleftrightarrow {CD}$, where $C=(3,-6)$, $D=(6,-2)$, and $B=(6,0)$? I can do this by using calculus but I think there has to be a nicer/shorter solution. Thanks in advance.

Edit: I found that $CE$ has distance $3$, so maybe one could use the theorem of Thales to obtain a right angle in the triangle $(0,-6), E,(6,-6)$ and use this?

  • $\begingroup$ maybe you could find the equation of the line and use it together with the circle to show that there is only one pair of solutions. $\endgroup$ – The Integrator Jun 3 '18 at 19:43
  • $\begingroup$ I'm not sure I understand. Do you mean to ask if it can be proved that the line through $CD$ intersects the circle? Or do you mean to ask if it can be proved that the line intersects the circle specifically in between the points $C$ and $D$ (ie at $B$)? $\endgroup$ – Ishan Jun 3 '18 at 19:44
  • $\begingroup$ @Ishan I mean the first option. $\endgroup$ – Luke Mathwalker Jun 3 '18 at 19:50

11 Answers 11


Extrapolate the line until it it intersects both coordinate axes. The points of intersection are then $(15/2,0)$ and $(0,-10)$.

The line and the exes now make a right triangle whose legs are $15/2$ and $10$. Now, apply the Pythagorean Theorem, find that the hypoteneuse is $25/2$ (proportional to a 3-4-5 right triangle).

In a right triangle, the area is half the product of the legs and also half the product of the hypoteneuse and the altitude to that hypoteneuse. So the products must be equal and we must have $(10)(15/2)=(25/2)x$ where $x$ is the altitude to the hypoteneuse. Then $x=6$ matching the radius of the given circle; the line is tangent to the circle.

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    $\begingroup$ This is the sort of proof I was looking for. Thanks! $\endgroup$ – Luke Mathwalker Jun 3 '18 at 20:35
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    $\begingroup$ @LukeMathwalker You can wait a little longer, but considering what you said in your comment, if you don't see an even better answer by tomorrow I would recommend clicking the checkmark to accept this one. $\endgroup$ – David K Jun 4 '18 at 2:32

The circle intersects the line if the radius of the circle is at least as large as the distance from the circle's center to the line.

We measure the distance from a point to a line in a direction perpendicular to the line. Seeing that the slope of the line from $C$ to $D$ is $\frac43,$ we want a line from the center of the circle whose slope is $-\frac34,$ that is, the line $y = -\frac34 x.$

Now we lay out the radius of the circle along that line. The radius is $6.$ Since the hypotenuse of a triangle with legs $3$ and $4$ is $5$, we can scale up one such triangle by the factor $\frac65$ to find the desired point on the line $y = -\frac34 x.$ That triangle has legs $\frac{18}{5}$ and $\frac{24}{5},$ and hypotenuse $6,$ which means the point we are seeking is $\left(\frac{24}{5}, -\frac{18}{5} \right).$

If that point is on the line or on the opposite side of the line from the center of the circle, then the circle intersects the line.

As it turns out, the point $\left(\frac{24}{5}, -\frac{18}{5} \right)$ is exactly on the line through $C$ and $D,$ which you can show using geometry without calculus.

A possibly more obvious approach is to find the equation of the line and then compute its distance from the center of the circle.

There are standard formulas for that sort of thing, but if you do not recall the distance formula it is easy enough to derive by building a right triangle with its right angle at the center of the circle, its hypotenuse on the line, and its legs parallel to the axes. (In this case, in fact, its legs are on the axes.)

Find the area of the triangle two ways: using the lengths of the legs, and using the length of the hypotenuse and the altitude from the right-angle vertex. The areas must be the same, of course, since it's the same triangle. Solve for the altitude, which is the distance to the line.

And yet another alternative:

Find the area of the triangle $\triangle ACD.$ Using this area and the fact that $CD = 5,$ find the altitude of $\triangle ACD$ from vertex $A.$ If the altitude is no greater than $6,$ the circle intersects the line.

  • $\begingroup$ How do you find the area of the triangle $\Delta ACD$? $\endgroup$ – Luke Mathwalker Jun 3 '18 at 20:41
  • $\begingroup$ Tri(AXY) - Tri(AXC) - Tri(ABY) = Tri(ABC). Work out the whole area enclosed by the axes and the line, then take away the two bits you don't want. The altitude and base for each is straightforward, and the base of Tri (ABC) is not much harder than Pythagoras. $\endgroup$ – Nij Jun 3 '18 at 22:35

The line through C and D has equation $$ 4x-3y-30=0 $$ The distance from the line to the point $(0,0)$ is given by the formula $$ d=\frac{|4(0)-3(0)-30|}{\sqrt{4^2 + (-3)^2}}$$ If $d \le 6$ we have intersection.

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    $\begingroup$ And if d=6 (as it is) we have a tangent. $\endgroup$ – marty cohen Jun 3 '18 at 20:56

The circle has equaton

  • $x^2+y^2=36$

the line hase equation

  • $y+6=\frac{-2+6}{6-3}(x-3)\implies y=\frac43x-10$

then we can check the intersection by solving the system of these two equations, that is

$$x^2+\left(\frac43x-10\right)^2=36\implies x^2+\frac{16}{9}x^2-\frac{80}{3}x+100-36=0 \\\implies25x^2-240x+576=0\implies(5x-24)^2=0 \implies x=\frac{24}{5} \quad y=-\frac{18}{5}$$

and since we have found an unique solution th eline is tangent to the circle at that point.

  • $\begingroup$ This was what I meant by "using calculus". I thought weither there is a more elegant way. But thanks nevertheless. $\endgroup$ – Luke Mathwalker Jun 3 '18 at 19:49
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    $\begingroup$ @LukeMathwalker That is not calculus. That is just analytic geometry and algebra. $\endgroup$ – David K Jun 3 '18 at 19:49
  • $\begingroup$ @LukeMathwalker calculus would be calculating the slopes of the curves and comparing them. this is geometry with some algebra in it $\endgroup$ – The Integrator Jun 3 '18 at 19:50
  • $\begingroup$ @LukeMathwalker Yes as DavidK just clarified that's not calculus but analytic geometry and algebra, we don't need derivatives by that approach. $\endgroup$ – gimusi Jun 3 '18 at 19:50
  • $\begingroup$ Probably there is a more elegant way. $\endgroup$ – Michael Hardy Jun 3 '18 at 19:50

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Note that $\operatorname{Area}(\Delta DFE)=\operatorname{Area}(\square ABCD)-\operatorname{Area}(\Delta DCF)-\operatorname{Area}(\Delta EFB)-\operatorname{Area}(\Delta DEA)$ which gives, since $\overline{CF}=2,\overline{AE}=3,\overline{AB}=\overline{BC}=\overline{CD}=\overline{DA}=6$: $$\operatorname{Area}(\Delta DFE)=36-\frac{1}{2}\cdot 6\cdot 2-\frac{1}{2}\cdot 3\cdot 4-\frac{1}{2}\cdot 3\cdot 6=15.$$ But $15=\operatorname{Area}(\Delta DFE)=\frac{1}{2}\overline{EF}\cdot\overline{DG}$. By the Pythagoras theorem $\overline{EF}=5$ so that $$\overline{DG}=\frac{2\cdot 15}{5}=6$$ which was to be proved.

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    $\begingroup$ I think this is a very logical way to approach the problem. $\endgroup$ – David K Jun 4 '18 at 2:29

Since the slope of the line is $\frac43$, the slope of the normal from A to the line is $-\frac34$.

The equation of this line is $y = -3x/4$. Since the equation of the circle is $x^2+y^2=36$, the line intersects the circle when $36 =x^2+(3x/4)^2 =x^2(1+9/16) =25x^2/16 $ so $6 = 5x/4$ or $x = 24/5$ and $y = -(24/5)(3/4) =-18/5 $.

This is on the line if $\dfrac{-2-(-6)}{6-3} =\dfrac{-18/5-(-6)}{24/5-3} $ or $\dfrac{4}{3} =\dfrac{12/5}{9/5} $ which is true.


Yes, you set the equation for the circle equal to the equation of the line and solve for x. If there is only one solution, the line is tangent and if two solutions the line cuts through the circle.

$\sqrt{6^2 - x^2} = \frac{4}{3}x - 10$

$36 - x^2 = \frac{16}{9}x^2 - \frac{80}{3}x + 100$

$\frac{25}{9}x^2 - \frac{80}{3}x + 64 = 0$

$(\frac{5}{3}x - 8)^2 = 0$

$\frac{5}{3}x = 8$

$x = 4.8$

  • $\begingroup$ I would avoid using the square root here. Taking the square root of both sides of an equation can eliminate solutions, and squaring both sides of an equation can introduce extra solutions. $\endgroup$ – aschepler Jun 5 '18 at 1:35
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    $\begingroup$ It's a perfect square so both roots are $(\frac{5}{3}x - 8) = 0$, I just didn't repeat myself, but thanks. $\endgroup$ – Phil H Jun 5 '18 at 1:57

Given $p = (x,y), p_1 = (3,-6), p_2 = (6,-2)$ we have a circle and a line

$$ C\to \|p\|^2 = r^2\\ L\to p = p_1+\lambda (p_2-p_1) $$

their intersection is calculated by solving

$$ C\cap L \to \|p_1+\lambda(p_2-p_1)\|^2= r^2 $$


$$ \lambda^2\|p_2-p_1\|^2+2\lambda(p_2-p_1)\cdot p_1 +\|p_1\|^2-r^2=0 $$


$$ \lambda = \frac{-2(p_2-p_1)\cdot p_1 \pm \sqrt{4((p_2-p_1)\cdot p_1)^2-4\|p_2-p_1\|^2(\|p_1\|^2-r^2)}}{2\|p_2-p_1\|^2} $$

but if $L$ is tangent to $C$ then

$$ 4((p_2-p_1)\cdot p_1)^2-4\|p_2-p_1\|^2(\|p_1\|^2-r^2)=0 $$


$$ ((p_2-p_1)\cdot p_1)^2-\|p_2-p_1\|^2(\|p_1\|^2-r^2)=0 $$


A strict answer to your question is: NOPE, there is no such way.

'Geometric way' means 'by construction' which is essentially 'by drawing' – and you can find an 'obvious' intersection by drawing, at best. And 'obvious' means you can demonstrate the possible drawing imperfections do not alter the result significantly.

However, you can't prove tangency, as this can significantly change (i.e., appear or disappear) on arbitrarily small errors in drawing. Tangency can be proven algebraically only (calculus not needed).

  • $\begingroup$ No, "construction" does not mean "drawing". $\endgroup$ – Acccumulation Jun 4 '18 at 22:00

Your original question asked whether there is a way to prove that the circle intersects the line, but you seem to be asking whether it's tangent. We can calculate the distances $Ac = 3\sqrt{5}, AD=2\sqrt{10}, CD = 5$.

Let's choose say there's a point $E$ such that $AE = 6$ (so it's on the circle) and ADF and ACF are right triangles. So $(CE)^2=(Ac)^2-(AF)^2=45-36=9$ and $(DE)^2=(AD)^2-(AF)^2=40-36=4$. Thus, $CE = 3, DE = 2$. The triangle inequality states that given any triangle, every side is less than the sum of the other two sides. Applying this to the "triangle" CED, we find that the side $CD$ is exactly equal to the sum of the other two sides, which means that it degenerate (not a true triangle); that is, the three point are all on the same line.

So $E$ is on both the circle and the line, and $AE$ is perpendicular to the line. Thus, $E$ is a point of tangency.


Here is a very simple proof, using basic geometry only.

The line segment CD (3,-6) -> (6,-2) is part of a line with equation y = 4x/3 - 10 with gradient 4/3 (basic maths). A normal to this line (perpendicular) would have a gradient -1/gradient = -3/4 (basic gradients). So perpendiculars to CD all have the form 'y = -3x/4 + c` where c is a real constant.

The closest that any line comes to the centre of the circle is the perpendicular distance from that line to the centre of the circle. Let's call that closest point (X,Y). Then:

  • Y = 4X/3 - 10 (it's on CD)
  • Y = -3X/4 + c for some c (it's on one specific normal, since precisely one normal intersects CD at any given point, and all normals have an equation like this)
  • 0 = -3(0)/4 + c = 0 + c (the centre of the circle at (0,0) is also on the same normal as (X,Y) so it obeys the same equation y=-3x/4+c for the normal as (X,Y) does, and for the same value of c)

From (3) we see that c = 0. Therefore we have Y = 4X/3 - 10 and Y = -3X/4. Multiplying both by 12 for simplicity we get: 12Y = 16X -120 and 12Y = -9X. Therefore 16X -120 = -9X, so 25X = 120 and X =120/25 = 4.8, and Y = -3X/4 = -3.6. So (X,Y) is (4.8, -3.6).

(X,Y) was defined as the closest that CD comes to the centre of the circle, and the distance from (4.8,-3.6) to (0,0) is given by Pythagorus: sqrt(4.8^2 +3.6^2) = 6

Therefore the closest CD gets to the centre of the circle is 6 units, at the point (4.8,-3.6). But as the circle is also exactly 6 units radius (because B is at (6,0)), the closest point must also be on the circumference of the circle.

But a line whose closest point to the centre of a circle is exactly on its circumference (but doesn't get closer than its circumference) must be a tangent to the circle, otherwise it would pass through the circumference and there would be an even closer point somewhere else.

So CD must be tangential to the circle and just touch (but not cross) it.


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