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Converters Constants Calculator Analytical Geometry
Circle VectorsStraight Line ParabolaEllipse Hyperbola
Coordinate Geometry [Triangle] Coordinate Geometry [Quadrilateral]
Operations:
Intersection of two circles
Intersection of a circle and a straight line
Equation of circumcircle of a triangle
Image of a circle
Equation of incircle of a triangle
Equation of excircles of a triangle
Equation of common chord of two circles
Parametric Equation of Circle
Chord of Contact of Circle

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OPERATIONS:

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Intersection of two circles
Intersection of a circle and a straight line
Equation of circumcircle of a triangle
Image of a circle
Equation of incircle of a triangle
Equation of excircles of a triangle
Equation of common chord of two circles
Parametric Equation of Circle
Chord of Contact of Circle
General Info
In Polar Coordinates
From two end-points of the diameter
Passing through three points
Touching both the axes
Passing through the origin
Touching a tangent and passing through a point
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

dialogs

Equation of a chord bisected at a point in a circle

Input the equations of the circle and the bisecting point of the chord.

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.



\(x^2+y^2+2gx+2fy+c=0\)

The circle has center \((-g, -f)\) and radius \(\sqrt{g^2+f^2-c}\).

Create the only circle:


Input the coordinate of the points on which the chord of the circle is bisected:








Theory [Equation of a chord bisected at a point in a circle]


Consider that the equation of the circle is: $$x^2+y^2+2gx+2fy+c=0$$ The circle has center \((-g,-f)\). Since the chord gets bisected at point \((p,q)\), then the equation of the chord must be perpendicular to the line joining the center of the circle and the bisection point of the chord. Therefore, we get the line: $$\frac{x+g}{-g-p}=\frac{y+f}{-f-q}$$ $$\Rightarrow x(f+q)-y(g+p)+gq-pf=0$$ A line perpendicular to this one passing through \((p,q)\) is the equation of our chord, therefore: $$x(g+p)+y(f+q)+gp+p^2+fq+q^2=0$$ Another shorter method is to, consider the formula \(T=g(p,q)\), where \(T\) is the equation of tangent at the point \((p,q)\) of the circle, which gives us: $$px+qy+x(g+p)+y(f+q)=p^2+q^2+2pg+2fq$$ where we have found \(T\), the equation of tangent at point \((p,q)\) of the circle by substituting \(x^2=xp\), \(y^2=yq\), \(2x=x+p\) and \(2y=y+q\).

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