orthogonal circles

Two circles intersecting orthogonally (http://planetmath.org/ConvexAngle) are orthogonal curves and called of each other.

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Since the tangent of circle is perpendicular to the radius drawn to the tangency point, the both radii of two orthogonal circles drawn to the point of intersection and the line segment connecting the centres form a right triangle. If  $(x-a_{1})^{2}+(y-b_{1})^{2}=r_{1}^{2}$  and  $(x-a_{2})^{2}+(y-b_{2})^{2}=r_{2}^{2}$  are the equations of the circles, then, by Pythagorean theorem,

 $\displaystyle r_{1}^{2}+r_{2}^{2}=(a_{2}-a_{1})^{2}+(b_{2}-b_{1})^{2}$ (1)

is the condition of the orthogonality (http://planetmath.org/OrthogonalCurves) of those circles.

The equation (1) tells that the centre of one circle is always outside its orthogonal circle.  If  $(x_{0},\,y_{0})$  is an arbitrary point outside the circle  $(x-a)^{2}+(y-b)^{2}=r^{2}$,  one can always draw with that point as centre the orthogonal circle of this circle:  its radius is the limited tangent from  $(x_{0},\,y_{0})$  to the given circle. The square (http://planetmath.org/SquareOfANumber) of the limited tangent is equal to the power of the point with respect to the circle and thus  $(x_{0}-a)^{2}+(y_{0}-b)^{2}-r^{2}$.  Accordingly, the equation of the orthogonal circle is

 $(x-x_{0})^{2}+(y-y_{0})^{2}=(x_{0}-a)^{2}+(y_{0}-b)^{2}-r^{2}.$

One of two orthogonal (http://planetmath.org/Orthogonal) circles harmonically any diameter of the other circle.

Title orthogonal circles OrthogonalCircles 2013-03-22 17:41:32 2013-03-22 17:41:32 pahio (2872) pahio (2872) 7 pahio (2872) Topic msc 51N20 PerpendicularityInEuclideanPlane orthogonal circle