## You are here

Homeorthogonal circles

## Primary tabs

# orthogonal circles

Two circles intersecting orthogonally are orthogonal curves and called orthogonal circles of each other.

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 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 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 circles divides harmonically any diameter of the other circle.

## Mathematics Subject Classification

51N20*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections