I have a problem in finding the radius and the center of a circle with the following conditions

  • Circle tangential to line $g_1$ and going through known point $P_1$
  • Circle is tangential to $g_2$

enter image description here

The following things I need to find:

  • radius of circle $(r)$
  • center of circle $(x_m, y_m)$
  • coordinates of $P_2 (x_2,y_2)$

So far I know that because of the horizontal line the middle point of the circle need to be $x_m = x_1$. I then tried to continue with the equation of a circle but I have still $2$ unknowns ($y_m$ and $r$) because I do not know how to use the constraint of $g_2$.

Does anybody have a hint how to solve this problem?

Edit #1: I tried to understand the "Special cases of Apollonius' problem" with two lines and one point. Here is the current progress: But I do not know how to continue since becaues P1 is already a point on the circle/the tangent.

enter image description here

  • $\begingroup$ First: find intersection P of g1 and g2. Second: find point P2 of g2 such that PP1=PP2. There are two such points, there are two circles, satisfying problem statement. Third: find equations of lines g3 and g4 perpendicular to g1 and g2 and passing through P1 and P2. Fourth: find intersection of g3 and g4. This is center O. Fifth: find distance OP1, which is radius. $\endgroup$ 2 days ago

2 Answers 2


Let $P_1 = (x_1, y_1)$. You can first find the intersection of $g_1$ and $g_2$.

Let the equation of $g_2$ be

$$ A x + B y + C = 0 $$

Set $ y = y_1$ and solve for $x$, this will give you,

$$ x = - \dfrac{B y_1 + C}{A} $$

Note that $A$ cannot be zero, because $g_2$ is not horizontal. If it is horizontal, then the problem becomes trivial to solve.

So now the intersection point between $g_1$ and $g_2$ is

$P_3 = (x_3, y_3) = (- \dfrac{B y_1 + C}{A} , y_1 ) $

Find the signed distance between $P_1$ and $P_2$:

$ d = x_1 - x_3 $

Next, find the angle between $g_1$ and $g_2$. There are two possible angles. The unit normal to $g_1$ is $u_1 = \langle 0, 1 \rangle$, and the unit normal to $g_2$ is $ u_2 = \langle \dfrac{A}{\sqrt{A^2 + B^2}}, \dfrac{B}{\sqrt{A^2 + B^2}} \rangle $. To get the angle right, we have to use a combination of dot product and cross product of $u_1$ and $u2$.

$ \cos(\phi_1) = u_1 \cdot u_2 $

$ \sin(\phi_1) = \bigg[ u_1 \times u_2 \bigg]_z $

And $ \phi_1 = \text{Atan2}(\cos(\phi_1), \sin(\phi_1) ) $

Therefore, the two possible angles between $g_1$ and $g_2$ are

$ \phi_1 $ and $ \phi_2 = \pi - \cos^{-1} ( u_1 \cdot u_2 ) $

From this, it follows that the possible signed radii are

$ r_1 = d \tan \bigg( \dfrac{\phi_1}{2} \bigg) $


$r_2 = - d \tan \bigg( \dfrac{\phi_2}{2} \bigg) $

And the center will be at

$ C_i = P_1 + (0, r_i) $

As an example, suppose $g_1$ is the line $y = 10$, and $g_2$ is the line $ y = \sqrt{3} (x + 15) $

First find the intersection point between the two lines

$ 10 = \sqrt{3} (x + 15) $ implies $ x_3 = -15 + \dfrac{10}{\sqrt{3}} $

Suppose $P_1 = (10, 0) $ , then $ d = x_1 - x_3 = 25 - \dfrac{10}{\sqrt{3}} $

$ u_1 = \langle 0, 1 \rangle$, $ u_2 = \langle - \dfrac{\sqrt{3}}{2} , \dfrac{1}{2} \rangle $

From this

$ \cos(\phi_1) = \dfrac{1}{2} $

$ \sin(\phi_1) = \dfrac{\sqrt{3}}{2} $

Therefore, $ \phi_1 = \dfrac{\pi}{3} $

And $ \phi_2 = \pi - \phi_1 = \dfrac{2 \pi}{3} $

From this,

$r_1 = d \tan(\dfrac{\pi}{6} ) = (25 - \dfrac{10}{\sqrt{3}}) (\dfrac{1}{\sqrt{3}} ) = \dfrac{1}{3} ( 25 \sqrt{3} - 10 ) $


$r_2 = - d \tan(\dfrac{\pi}{3} ) = - (25 \sqrt{3} - 10 ) $

Here is additional figures obtained from this algorithm.

enter image description here

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  • $\begingroup$ Wow I'm really impressed. Thank you for this great solution. One question to that. Why there is a minus sign at r2? In my understanding a radius is always unsigned? $\endgroup$
    – mk3
    2 days ago
  • $\begingroup$ It is so you can add the signed radius to the point $P_1$ to obtain the center $C$ as follows $C_i = P_1 + (0, r_i) $ $\endgroup$ 2 days ago
  • $\begingroup$ Thanks for the explanation. Very nice solution $\endgroup$
    – mk3
    2 days ago
  • $\begingroup$ Just an additional question. Is this method also applicable if I have still two lines and not only P1 on g1 but also P2 on g2 is fixed? Or is this problem then overconstrained? Because now P2 is also a solution but what if it is an additional constraint? $\endgroup$
    – mk3
    2 days ago
  • $\begingroup$ Yes. In this case, the problem will be over-constrained. $\endgroup$ 2 days ago

If a circle is tangent to two lines, its center must be on one of the two angular bisectors of these two lines.

If a circle is touching a line in a given point, the center must be on the orthogonal line through that point, as you already found out.

This is a special case of Apollonius' problem, namely the one with two lines and one point, made even more special by the point lying on one of the lines.

  • $\begingroup$ Thank you for your hint. I tried to understand what I can find there but still do not know how to continue from the line connecting P1 and P2 on. The result of my understanding you can find at Edit #1. $\endgroup$
    – mk3
    2 days ago

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