# Circle with center point and tangential to lines

I have defined Points all points (3 blue, and one green). All points have the same distance to A point. Yellow lines are bisectors. I have equations of AB and AC with Ax + By +C = 0 form.

I need construct circle (tangential to lines) with green point center. Any hint or steps? BR

• If the centre is given, then it is possible that no circle can be tangential to both (extensions of) $AB$ and $AC$. Commented Dec 20, 2014 at 17:31

You should find two things:

1. Center of the circle: it's green points: $G=(x_1,y_1)$.
2. Radius of the circle: it's a distance from a point $G$ to a line $Ax+By+C=0$ (see here):

$$r=\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$$

Equation of the circle is:

$$(x-x_0)^2+(y-y_0)^2=r^2$$

• That is cool and simply solution but one problem is that i have all 4 points , and don't know which is this one between AB and AC Commented Dec 20, 2014 at 18:00
• One heuristic solution is: find projection of all 4 point onto lines $AB$ and $AC$ and checking which one lay between $A$ and $C$ on line $AC$ and $A$ and $B$ on $AB$ and then apply above solution to right point.
– agha
Commented Dec 20, 2014 at 18:28
• what do you mean by "projection of all 4 point onto lines AB and AC" ? How to do this? Commented Dec 20, 2014 at 18:31
• Note that line $-Bx+Ay+Bx_0-Ay_0=0$ is perpendicular to line $Ax+By+C=0$ and it passes through point $G=(x_0,y_0)$. Now find intersection line $Ax+By+C=0$ and $-Bx+Ay+Bx_0-Ay_0=0$. Do this with two lines: passes through $A$ and $C$ passes through $A$ and $B$. Check if this intersections lay between $A$ and $B$ and between $A$ and $C$ -if it's true it's "green" point.
– agha
Commented Dec 20, 2014 at 18:41
• are you sure this equation −Bx+Ay+Bx0−Ay0=0 ? Commented Dec 20, 2014 at 18:56

Construct a line through the green point perpendicular to one of the two lines. You now have the radius of your circle (distance from the green point to the intersection of that perpendicular line and the other line). Now draw the circle with that radius and you're done.

EDIT: I assumed the OP meant construction using a ruler and a compass. This can also be done analytically in a very similar way. The main difference being that the perpendicular line must be found analytically (I recommend using vectors).