Passing through Intersecting point of two circles and another point
Passing through Intersecting point of circle and a straight line and another point
Chord Bisected at a Point
Having a center and a tangent
Cartesian Coordinate to Polar Coordinate
Polar Coordinate to Cartesian Coordinate
dialogs
Polar Equation of a Circle from Cartesian Equation
Input the equation of the circle to find out its polar equation.
In case of inputting square roots, press the square root button and activate it. To insert fractions, insert it converting into decimals. For example, \(\frac{3}{4}\) is to be inputted as \(0.75\). But the output will be in decimals. Again, for working with more irrational numbers like \(2+\sqrt3\), you must input it in decimal form using full-stop \((.)\) as the decimal point.
\(x^2+y^2+2gx+2fy+c=0\)
\((x-a)^2+(y-b)^2=r^2\)
The circle has center \((-g, -f)\) and radius \(\sqrt{g^2+f^2-c}\).
Create the only circle:
The circle has center \((a, b)\) and radius \(r\).
Create the only circle:
dialogs
Cartesian Equation of a Circle from Polar Equation
Input the polar equation of the circle to find out its Cartesian equation.
In case of inputting square roots, press the square root button and activate it. To insert fractions, insert it converting into decimals. For example, \(\frac{3}{4}\) is to be inputted as \(0.75\). But the output will be in decimals. Again, for working with more irrational numbers like \(2+\sqrt3\), you must input it in decimal form using full-stop \((.)\) as the decimal point.
To input angles, input in radian format. To input angles in format of \(\frac{\pi}{4}\) press the button that has the symbol of \(\pi\) and activate it. As an example, \(\frac{3\pi}{4}\) might be inputted as \(0.75\) with the \(\pi\) button pressed.
\(r^2-2r(gcos\theta+fsin\theta)+c=0\)
\(r^2-2rr_1cos(\theta-\theta_1)+r_1^2=a^2\)
The circle has center \((-g, -f)\) in Cartesian Coordinates or $$\left(\sqrt{g^2+f^2},tan^{-1}\frac{f}{g}\right)$$ in Polar Coordinates and radius \(\sqrt{g^2+f^2-c}\).
Create the only circle:
The circle has center \((r_1,\theta_1)\) in Polar Coordinates and radius \(a\).
Create the only circle:
Theory [Equation of a Circle in Polar or Cartesian Coordinates]
A circle generally has the equation: $$x^2+y^2+2gx+2fy+c=0$$ We just substitute \(r^2=x^2+y^2\) and \(x=-\cos\theta\) alongwith \(y=-\sin\theta\). After transforming into polar coordinates, we can find the center of the circle in polar coordinate (a coordinate system that has \(r\) and \(\theta\) as the orthogonal axes), where center is \(r=\sqrt{g^2+f^2}\) and polar angle \(\theta=\tan^{-1}\left(\frac{y}{x}\right)\). It is not mandatory to have \(x=-\cos\theta\) and \(y=-\sin\theta\) set as they are, you can shift them and recalculate the polar angle in the different way. If the circle has the format $$\left(x-a\right)^2+\left(y-b\right)^2=r^2$$ then convert it into \(x^2+y^2+2gx+2fy+c=0\) format. To convert from polar coordinate to cartesian coordinate, substitute \(x^2+y^2=r^2\) and \(x=-\cos\theta\) alongwith \(y=-\sin\theta\). The parameter \(c\) remains unchanged in every case. In case of a circle having center \(r_1,\theta_1\) and a radius \(a\), we need to consider that the center in Cartesian Coordinate: $$-g=r_1\cos\theta_1$$ $$-f=r_1\sin\theta_1$$ where we have \((-g,-f)\) as the center of the circle. Now the radius has the same equation as before: $$a^2=g^2+f^2-c$$ $$\Rightarrow a^2=r_1^2-c$$ $$\Rightarrow c=r_1^2-a^2$$
