How to find a tangent circle is inside a quadrant ellipse, the circle is tangent to +x axis, +y axis and the ellipse. half -axis of ellips are a=2 and b=1. what is the diameter of tangent circle (r=?). I solved this problem by iteratively. But I want to solve this problem with closed formula . can you help me? thanks.
MATLAB: How to find a tangent circle is inside a quadrant ellipse
circleellipseMATLABtangent
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Try this:
x = linspace(-2, 2, 500);y = x .^ 2 + 2;plot(x, y, 'b-', 'LineWidth', 2);grid on;fontSize = 20;xlabel('X', 'FontSize', fontSize);ylabel('Y', 'FontSize', fontSize);ylim([0, 7]);% Pick a point at -1 (or wherever) and plot the tangent
xTangent = -1;% Equation of the tangent is y = slope*x + offset
% From the equation y=x^2 we know the slope is 2*x.
% Find the slope at x = xTangent.
slope = 2 * xTangent;% We want to plot the tangent line where it just touches the curve,
% so we need to know the y value at xTangent.
yTangent = xTangent .^ 2 + 2; % Y value of curve at x = xTangent.
hold on;plot(xTangent, yTangent, 'r*', 'LineWidth', 2, 'MarkerSize', 10);% Use point slope formula of a line to get equation for y
% y-y0 = slope*(x-x0).
% y = slope * (x - xTangent) + yOffset
yTangentLine = slope * (x - xTangent) + yTangent;plot(x, yTangentLine, 'b-', 'LineWidth', 2);
Hi Manoj,
linecirc gives the intersection points of a line and a circle.
Syntax:
[xout,yout] = linecirc(slope,intercpt,centerx,centery,radius);This function finds the points of intersection given a circle defined by a center and radius in x-y coordinates, and a line defined by slope and y-intercept, or a slope of “inf” and an x-intercept. Two points are returned. When the objects do not intersect, NaNs are returned.
When the line is tangent to the circle, two identical points are returned. All inputs must be scalars.
Example:
[xout, yout] = linecirc(-1,3,0,0,5);This will give the intersection point of line 
and circle 
.
and circle .
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