Common chord of two circles calculator

To use the calculator, enter the x and y coordinates of a center and radius of each circle.

A bit of theory can be found below the calculator.

Calculation precision

Digits after the decimal point: 2

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Circles intersection

The task is relatively easy, but we should take into account the edge cases – therefore we should start by calculating the cartesian distance d between two center points, and checking for edge cases by comparing d with radiuses r1 and r2.

Here are the possible cases (distance between centers is shown in red):

Case Description Rule

	Trivial case: the circles are coincident (or it is the same circle)
	The circles are separate
	One circle is contained within the other
	Two intersection points	You have one or two intersection points if all rules for the edge cases above are not applied
	One intersection point	Trivial case of two intersection points

So, if it is not an edge case, to find the two intersection points, the calculator uses the following formulas (mostly deduced with Pythagorean theorem), illustrated with the graph below:

Two intersection points

The first calculator finds the segment a

and then the segment h

To find point P3, the calculator uses the following formula (in vector form):

And finally, to get a pair of points in case of two points intersecting, the calculator uses these equations: First point:

Second point:

Note the opposite signs before the second addend

For more information, you can refer to Circle-Circle Intersection and Circles and spheres

The common chord of the two circles c1 and c2 is 3.8 cm long. This chord forms an angle of 47° with the radius r1 of the circle c1 and an angle of 24° 30´ with the radius r2 of the circle c2. Calculate both radii and the distance between the two centers of the circles.

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We encourage you to watch this tutorial video on this math problem: video1

Answer

Verified

Above is the diagram of the two circles having a common chord and we know that it's perpendicular to the line joining P and Q that is PQ is perpendicular. We need to find the length of chord AB.As we know PQ is perpendicular bisector of AB so AO = OB……(1)So, we will apply Pythagoras theorem to get the value of PO to get the value of AO to get the right answer.It is given that PQ = 25cmLet PO = x, then QO = 25 – x.Applying Pythagoras in Triangle AOP we get,$   \Rightarrow A{O^2} = A{P^2} - O{P^2} \\   \Rightarrow A{O^2} = {15^2} - {x^2}.........(2) \\ $Again applying Pythagoras in triangle AOQ we get,$   \Rightarrow A{O^2} = A{Q^2} - O{Q^2} \\   \Rightarrow A{O^2} = {15^2} - {\left( {25 - x} \right)^2}.........(3) \\ $From (2) and (3) we can say that,$  {20^2} - {\left( {25 - x} \right)^2} = {15^2} - {x^2} \\  400 - 625 - {x^2} + 50x = 225 - {x^2} \\  50x = 450 \\  x = 9 \\ $So, PO = 9cm then OQ = 16cmNow we can apply the Pythagoras theorem in any of the triangle AOP or AOQ to get the value of AO.So, we do,$ \Rightarrow A{O^2} = A{P^2} - O{P^2}$Putting the values we get,\[   \Rightarrow A{O^2} = {15^2} - {9^2} \\   \Rightarrow A{O^2} = 225 - 81 \\   \Rightarrow A{O^2} = 144 \\   \Rightarrow AO = 12cm \\ \]So, we get AO = 12cm we know that AB = 2AO from (1)So, the length of the chord AB = 2(12) = 24cm.

So, the correct answer is “Option A”.

Note: In this problem we have used various properties of a circle like the common chord is perpendicularly bisected by the line joining the centers of the two circles and then we have applied the Pythagoras theorem in the various triangles formed in the figure to get the right answer. Drawing the figure will help you a lot in such types of problems.

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If we know the radii of two intersecting circles, and how far apart their centers are, we can calculate the length of the common chord.

Problem

Circles O and Q intersect at points A and B. The radius of circle O is 16, and the radius of circle Q is 9. Line OQ connects the centers of the two circles and is 20 units long. Find the length of the common chord AB.

Strategy

We know that line OQ is the perpendicular bisector of the common chord AB. And we are also given the lengths of the radii, so we probably need to use that. Let's draw these radii:

Now, we have two triangles, △AOC and △AQC. They are both right triangles (since OQ is perpendicular to AB), and both have the same height, h. If the base of one of these is x units long, the other base is 20-x, as OQ is 20 units long.

Now, using the Pythagorean Theorem and some basic algebra, we can solve the following system of equations for x:

(1) h2+(20-x)2=162 (in right triangle △AOC)
(2) h2+x2=92 (in right triangle △AQC)

And once we find x, we substitute it in one of the above equations to find h. AB is 2h, since OQ is a bisector of AB.

Solution

(1) AB⊥OQ //Line connecting centers is perpendicular to common chord(2) △AOC, △AQC are right triangles

(3) h2+(20-x)2=162 //(2), Pythagorean theorem

(4) h2+x2=92 //(2), Pythagorean theorem(5) 400-40x = 256-81 //Subtract (4) from (3)(6) x=225/40=5.625

(7) h2=81-31.64

(8) h=√49.36=7.03

(9) AB=14.06