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**Contents:**

The circle also passes through R. You may investigate the case for which the absolute value is added:. Let S x, y be any point on the circle.

Gradient of SQ. Gradient of SR. Let b be the angle QSR. We do not include the absolute value in the formula.

The order of m is therefore important. Equate 1 and 2 , we get. Simplify, we get the equation of circle:. Use the Angles in the same segment. Use Opposite angles of cyclic quad. The required circle is of the form:.

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Similarly for Q and R. Therefore the circle passes through P, Q, R. Using Laplace expansion we get:. You should know how to break down a higher order determinant. Eliminating F and get a simultaneous equation in D and E. Solve D and E and finally get F. Find the equation of the circle passing through the points P 2,1 , Q 0,5 , R -1,2 Method 3 The perpendicular bisectors of two chords meet at the centre.

Find the equation of the circle passing through the points P -6,5 , Q -3,-4 , R 2,1 Method 4 Converse of Angle in semi-circle. Find the equation of the circle passing through the points P -6,5 , Q -3,-4 , R 2,1 Method 5 System of circles. Find the equation of the circle passing through the points P -6,5 , Q -3,-4 , R 2,1 Method 6 Angles in same segment. Find the equation of the circle passing through the points P 2,1 , Q 0,5 , R -1,2.

Method 1. Substitute points in the general form of circle and solve coefficients.

Method 2. Use Centre and Radius Form of the circle. This is the given equation.

Move the loose number over to the other side. Group the x -stuff and y -stuff together. Divide off by whatever is multiplied on the squared terms. Take the coefficient on the x -term, multiply by one-half, square, and add inside the x -stuff and also to the other side.

Do the same with the y -term. Convert the left-hand side to squared form, and simplify the right-hand side. If necessary, fiddle with signs and exponents to make your equation match the circle equation's format.

Read off the answer. Stapel, Elizabeth.

Accessed [Date] [Month] Study Skills Survey. When a circle is inverted in a circle it is inverted to either a circle or a line. The same holds when a line is inverted in a circle, it is inverted to either a circle or a line. It is convenient not to distinguish between a circle and a line, and instead use a so-called generalized circle , i.

An angle at a point on a circle is measured from the tangent through that point. There are always two supplementary angles between two generalized lines. When you invert three points in a circle, their orientation is reversed, just as when reflecting in a line. We will not prove the theorem for all cases but sketch out the proof for the case where the angle is formed by two circles.

This is shown by using the same steps as in Exercise 4. Other cases of angles between circles or lines can be proved in a similar way. Note that when two circles are reflected in a circle, the angles are preserved but since the orientation of points is reversed the orientation of angles is also reversed.

When forming the angles between two circles you can always do it so that exactly one of the supplementary angles is inside both circles. After inversion the two angles are reversed with regards to which angle that is inside both circles. As a consequence of angle preservation, objects that are tangent to each other, are inverted to objects that are also tangent to each other. A geodesic is the hyperbolic equivalence of a Euclidean line. The reason why perpendicular generalized arcs are used as hyperbolic lines to model hyperbolic geometry, is because of the properties of perpendicular circles when doing circle inversion.

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