Three tangent circle

[Sai Nikxith]

How to make a circle that is tangential to three straight lines?

I used Astute Graphics's "Tangent Circle Tool" to make it tangential to two of the three straight lines, but I want it to be tangential to all of the three straight lines. How to do it?

[Mylenium]

You could convert the lines to guides and just move and scale the circle. If you have Astute's Collider Scribe as well this would be even easier...

Mylenium

[Sai Nikxith]

Sorry, but can you please explain it with some images?

[Tom Winkelmann]

This helps to understand...

https://mathalino.com/reviewer/plane-geometry/centers-of-a-triangle

[Sai Nikxith]

I do understand the geometry here, and thanks for sharing it. But, I want to know how to do it in illustrator.

[CarlosCanto · Accepted solution]

source:

https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Elementary_Trigonometry_(Corral)/02%3A_G...

[Sai Nikxith]

Okkayy!! I've got to know how to do it now and I'm done with my work.

Just asking, is this the only way to do so, because I feel like we can do it more easily.

[Tom Winkelmann]

Some month/years ago, somebody ask the same question.

I didn't remember the users name, but somebody made a script for that.

[CarlosCanto]

Okkayy!! I've got to know how to do it now and I'm done with my work.

Just asking, is this the only way to do so, because I feel like we can do it more easily.

By @Sai Nikxith

it's definitely not the only way of doing it, is it worth your time (or anyone else's) to find an easier way?

[Jacob Bugge]

Sai Nikxith,

You can, Smart Guides being your friends, working on a copy of your three lines which are hidden beneath the copies:

1) Cut the lines at their intersections and delete the outer end parts so you work with a triangle, just keep them as separate lines;

2) In the stacking order (Layers panel) move the shortest line (lower rigth in your case) to the top if not already;

3) Click the top line to select it, then switch to the Rotate Tool, then Click one of the ends to set it as centre, then Alt/OptionClickDrag the other end to snap a copy to the other line from that corner, then delete the copy; now you have the full angle between the lines/sides;

4) Alt/OptionClick the same corner, then divide the angle by 2 (just insert /2 in the box) and Copy; this is your first angle bisector;

5) Repeat 3) and 4), only at the other end; this is your second angle bisector, and their intersection is the centre of the incircle;

6) Switch to the the Selection Tool and click the top line, then copy it and rotate the copy by 90 degrees, then deselect and ClickDrag it by the relevant end Anchor Point to snap to the centre, then cut it at the intersection with the top line and delete the outer end part; now you have one normal with the radius of the incircle;

7) Select the normal and copy its length from the Document Info, then create a circle with twice the length as W=H (just add *2 in the first box);

8) ClickDrag the circle by its centre to snap to the centre from 5), and delete the unwanted bits.

Edit: I had to leave for (quite) a while before finishing.

Hi Carlos.

How about establishing the radius/diameter of the circle, so it can be created? I create/use a normal in 6) and 7).

Edit edit: Here is a corresponding drawing, sides in black, angle bisectors in blue, normal in green, incircle in red; and I have improved the wording.

[CarlosCanto]

Hi Jacob, what about the radius/diameter? the value can be calculated mathematically, there's formula in the link I posted.

I didn't try your method yet but I'm sure it works.

[Jacob Bugge]

Carlos,

Whenever I can, for all sorts of things, I leave the calculations to Illy and just draw, so I never thought of looking into the link for formulas; now I have seen the tough stuff.

[CarlosCanto]

🙂

[tromboniator]

Jacob, I set out to find another way to do this construction. I tried this way and that way, and finally, AHA! I had it!

I had very cleverly rebuilt your method.

[Jacob Bugge]

Hi Peter.

Birds of a feather.

[m1b · Accepted solution]

I've written a script, based on maths I found (beyond my skills!) that draws the in-circle from three selected lines. (It can also draw the triangle made from the intersecting lines, but it's commented out for this post.)

/*

    Make InCircle From Intersecting Lines.js

    Draws a circle that fits inside the triangle made from three intersecting lines
    (Also can draw the triangle made from intersection of lines)

    by m1b (and vital contributors mentioned below!)
    posted here: https://community.adobe.com/t5/illustrator/three-tangent-circle/m-p/12204800

    I found the mathematical code here:
    https://dirask.com/posts/JavaScript-how-to-calculate-intersection-point-of-two-lines-for-given-4-points-VjvnAj
    https://www.geeksforgeeks.org/program-to-find-the-incenter-of-a-triangle/
    https://www.geeksforgeeks.org/program-to-find-the-radius-of-the-incircle-of-the-triangle/

    Notes: not tested much and very little sanity checking!

*/

(function () {

    // assume 3 line pathitems selected
    // note the lines are treated as essentially infinite length,
    // so they don't need to actually intersect

    try {
        if (selection.length != 3)
            throw 'Please select three lines and try again.';

        var line1 = selection[0];
        var line2 = selection[1];
        var line3 = selection[2];

        if (!(itemIsLine(line1) && itemIsLine(line2) && itemIsLine(line3)))
            throw 'Please select three lines and try again.'

        // this makes a simple 'triangle' object of where the lines intersect
        var myTriangle = triangleFromIntersectingLines(line1, line2, line3);

        // this draws the triangle
        // var myTrianglePathItem = makeTrianglePathItem(myTriangle);

        // this draws the in-circle of the triangle
        var myCirclePathItem = makeInCirclePathItem(myTriangle);

    } catch (error) {
        alert(error);
    }


    // functions
    function triangleFromIntersectingLines(line1, line2, line3) {
        var triangle = {};
        triangle.points = [
            pointOfIntersection(line1, line2),
            pointOfIntersection(line2, line3),
            pointOfIntersection(line3, line1)
        ];

        triangle.sideLengths = [
            distanceBetweenPoints(triangle.points[0], triangle.points[1]),
            distanceBetweenPoints(triangle.points[1], triangle.points[2]),
            distanceBetweenPoints(triangle.points[2], triangle.points[0])
        ];

        return triangle;
    }

    function makeTrianglePathItem(triangle) {
        var trianglePathItem = activeDocument.pathItems.add();
        trianglePathItem.setEntirePath(new Array(
            new Array(triangle.points[0][0], triangle.points[0][1]),
            new Array(triangle.points[1][0], triangle.points[1][1]),
            new Array(triangle.points[2][0], triangle.points[2][1])
        ));
        trianglePathItem.closed = true;

        return trianglePathItem;
    }


    function makeInCirclePathItem(triangle) {
        // triangle is an object with points and sidelength properties
        var circle = {};
        circle.radius = radiusOfInCircle(triangle);
        circle.center = calculateInCenter(triangle);

        // make the in-circle
        var circlePathItem = app.activeDocument.pathItems.ellipse(
            circle.center[1] + circle.radius, // top
            circle.center[0] - circle.radius, // left
            circle.radius * 2, // width
            circle.radius * 2 // height
        )

        return circlePathItem;
    }



    function pointOfIntersection(line1, line2) {
        // line1 and line2 are expected to be PathItems with 2 PathPoints
        if (line1 == undefined || line1.pathPoints == undefined || line1.pathPoints.length != 2)
            throw "pointOfIntersection error: line1 doesn't have two points";
        if (line2 == undefined || line2.pathPoints == undefined || line2.pathPoints.length != 2)
            throw "pointOfIntersection error: line2 doesn't have two points";
        var p1 = pointArrayToObject(line1.pathPoints[0].anchor),
            p2 = pointArrayToObject(line1.pathPoints[1].anchor),
            p3 = pointArrayToObject(line2.pathPoints[0].anchor),
            p4 = pointArrayToObject(line2.pathPoints[1].anchor),
            intersectionPoint = calculateIntersection(p1, p2, p3, p4);
        return pointObjectToArray(intersectionPoint);

        function pointArrayToObject(p) { return { x: p[0], y: p[1] } };
        function pointObjectToArray(p) { return [p.x, p.y] };
    }

    function calculateIntersection(p1, p2, p3, p4) {
        // p1, p2, p3 and p4 are expected to be objects with x and y values

        // lower part of intersection point formula
        var d1 = (p1.x - p2.x) * (p3.y - p4.y); // (x1 - x2) * (y3 - y4)
        var d2 = (p1.y - p2.y) * (p3.x - p4.x); // (y1 - y2) * (x3 - x4)
        var d = (d1) - (d2);

        if (d == 0) throw 'Number of intersection points is zero or infinity.';

        // upper part of intersection point formula
        var u1 = (p1.x * p2.y - p1.y * p2.x); // (x1 * y2 - y1 * x2)
        var u4 = (p3.x * p4.y - p3.y * p4.x); // (x3 * y4 - y3 * x4)

        var u2x = p3.x - p4.x; // (x3 - x4)
        var u3x = p1.x - p2.x; // (x1 - x2)
        var u2y = p3.y - p4.y; // (y3 - y4)
        var u3y = p1.y - p2.y; // (y1 - y2)

        // intersection point formula
        var px = (u1 * u2x - u3x * u4) / d;
        var py = (u1 * u2y - u3y * u4) / d;

        return { x: px, y: py };
    }


    function calculateInCenter(triangle) {
        var x1 = triangle.points[0][0],
            y1 = triangle.points[0][1],

            x2 = triangle.points[1][0],
            y2 = triangle.points[1][1],

            x3 = triangle.points[2][0],
            y3 = triangle.points[2][1],

            a = triangle.sideLengths[1],
            b = triangle.sideLengths[2],
            c = triangle.sideLengths[0];

        // Formula to calculate in-center
        var x = (a * x1 + b * x2 + c * x3) / (a + b + c);
        var y = (a * y1 + b * y2 + c * y3) / (a + b + c);

        return [x, y];
    }


    function radiusOfInCircle(triangle) {

        var a = triangle.sideLengths[0],
            b = triangle.sideLengths[1],
            c = triangle.sideLengths[2];

        // the sides cannot be negative
        if (a < 0 || b < 0 || c < 0)
            return -1;

        // semi-perimeter of the circle
        var p = (a + b + c) / 2;

        // area of the triangle
        var area = Math.sqrt(p * (p - a) * (p - b) * (p - c));

        // Radius of the incircle
        var radius = area / p;

        // Return the radius
        return radius;
    }


    function distanceBetweenPoints(p1, p2) {
        var a = p1[0] - p2[0];
        var b = p1[1] - p2[1];
        return Math.sqrt(a * a + b * b);
    }

    function itemIsLine(item) {
        return item != undefined && item.pathPoints != undefined && item.pathPoints.length === 2;
    }

})();

[CarlosCanto]

there you go, it couldn't get any easier for the OP

[Jacob Bugge]

Carlos,

Actually, working on it yesterday I thought of scripting, and I wondered whether the drawn way I was using could form the basis of, or inspiration for, a script, or the two ways are (too) complementary.

I have sometimes pondered over that in other cases, and wondered what you would say.

Besides, when I saw your way, I was surpsised that you chose drawing rather than scripting.

But I was too entangled in urgencies yesterday to think of asking those questions.

[CarlosCanto]

Jacob, I always think of writing a script for complex queries, but writing scripts takes time I didn't have. Also the OP wanted to know how do it, so I thought showing how do it manually was what he needed.

[Jacob Bugge]

Indeed, Carlos.

[Sai Nikxith]

Can't be easier than this.

THANKS A LOTTT!!!

[Kurt Gold]

Your script is really beautiful, m1b.

Thanks for making and sharing it.

[m1b]

Thanks Kurt, I appreciate the generous comment. It was fun script to make. Of course the real beauty is in the maths, for which I take no credit. 🙂

[Kurt Gold]

Well, you have taken the right and the beautiful math, m1b.

Therefore I dare say that you provided the script of the year so far (let's see what will be there in the forthcoming months).

[ajabon grinsmith · Accepted solution]

After selecting the three straight lines,

and "Divide" in Pathfinder,

all you have to do is drag the Live Corner to the center of the triangle.

I duplicated the triangle for illustration and used it as a guide.