How to draw three circles with centers and radii given?

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I have entered the data in this program and tried to draw three circles on same figure which would then give me their intersection in x and y but I can't cuz there is no command for circles. Kindly assist.
clc
clear all
close all
x1=2; y1=8; r1=10;
x2=5; y2=5; r2=12;
x3=4; y3=9; r3=9;
function circle(x,y,r);
h=plot(x
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

Error in connector.internal.fevalMatlab

Error in connector.internal.fevalJSON
drawCircle([x1;x2], r1, 'r');
drawCircle([x2;y2], r2, 'g');
drawCircle([x3;y3], r3, 'b');
x=((y2-y1)*((y2^2-y1^2)+(x2^2-x1^2)+(r1^2-r2^2))- (y1-y2)*((y3^2-y2^2)+(x3^2-x2^2)+(r2^2-r3^2)))/2*((x1-x2)*(y2-y3)-(x2-x3)*(y1-y2))
y=((x2-x3)*((x2^2-x1^2)+(y2^2-y1^2)+(r1^2-r2^2))- (x1-x2)*((x3^2-x2^2)+(y3^2-y2^2)+(r2^2-r3^2)))/2*((y1-y2)*(x2-x3)-(y2-y3)*(x1-x2))
plot(x,y)

Accepted Answer

Image Analyst
Image Analyst on 27 Jun 2022
"I can't cuz there is no command for circles" <== there is if you have the Image Processing Toolbox.
x1=2; y1=8; r1=10;
x2=5; y2=5; r2=12;
x3=4; y3=9; r3=9;
viscircles([x1, y1], r1, 'Color', 'r');
viscircles([x2, y2], r2, 'Color', 'g');
viscircles([x3, y3], r3, 'Color', 'b');
As far as determining intersection points, you'll have to use math for that. Or else use a distance formula on your (x,y) arrays.
  2 Comments
Curious
Curious on 13 Jul 2022
x1=2; y1=8; r1=10;
x2=5; y2=5; r2=12;
x3=4; y3=9; r3=9;
viscircles([x1, y1], r1, 'Color', 'r');
viscircles([x2, y2], r2, 'Color', 'g');
viscircles([x3, y3], r3, 'Color', 'b');
hold on
x=(((x2-x3)*((x2^2-x1^2)+(y2^2-y1^2)+(r1^2-r2^2))-(x1-x2)*((x3^2-x2^2)+(y3^2-y2^2)+(r2^2-r3^2))))/(2*((y1-y2)*(x2-x3)-(y2-y3)*(x1-x2)))
x = -14.8889
y=(((y2-y3)*((y2^2-y1^2)+(x2^2-x1^2)+(r1^2-r2^2))-(y1-y2)*((y3^2-y2^2)+(x3^2-x2^2)+(r2^2-r3^2))))/(2*((x1-x2)*(y2-y3)-(x2-x3)*(y1-y2)))
y = -4.5556
plot(x , y,'r*')
I have tried this code of yours with a mathematical formula for intersection point but it's not giving the intersection. Any thoughts?
Image Analyst
Image Analyst on 13 Jul 2022
You have to get the equation of a circle and set them equal to each other, like
(x-x1c).^2 + (y - y1c).^2 - R1^2 = (x-x2c).^2 + (y - y2c).^2 - R2^2
Then solve for x and y. It might be easier to do numerically than analytically but maybe not. You'd have to multiply it out, collect terms, etc. just like you'd solve any equation.

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More Answers (1)

KSSV
KSSV on 27 Jun 2022
x1=2; y1=8; r1=10;
x2=5; y2=5; r2=12;
x3=4; y3=9; r3=9;
th = linspace(0,2*pi) ;
x = cos(th) ; y = sin(th) ;
figure
hold on
plot(x1+r1*cos(th),y1+r2*sin(th)) ;
plot(x2+r2*cos(th),y3+r2*sin(th)) ;
plot(x3+r1*cos(th),y3+r3*sin(th)) ;
axis equal

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