Reflection Points for a Straight Line

where r is real and α is non-zero complex constant.

Inverse points of a circle -

Two points P and Q are said to be inverse w.r.t. a circle with centre 'O' and radius P if

(i) The points O, P, Q are collinear and P, Q are on the same side of O.

(ii) OP.OQ = P²

Now, as depicted in the figure above, (OP)(OP’) = r². Various conclusions follow form this equation like:

• If (OP) (OP’) = r², then (OP’)(OP) = r².

This means that if point P is inverse of point P’, then the point P’ is the inverse of point P.

• In case OP < r, then OP’ > r

If a particular point lies in the interior of the circle of inversion, then its inverse will lie in the exterior.

• In case OP > r, then OP’ < r

This case is quite opposite to the previous case. If a particular point lies in the exterior of the circle of inversion, then its inverse will lie in the interior.

• In case OP = r, then OP’ = r

Now, if the point lies on the circle of inversion itself then so does its inverse. Hence, the point in such a case may be regarded to be invariant. This means that the point is its own inverse.  

Remark: Mathematically, the inverse points may also be defined as:

Two points z₁ and z₂ will be the inverse points w.r.t. the circle

The concept of inverse points may also be considered in context of an inversion sphere. The inversion sphere may also be regarded as the extension of the geometric inversion form two dimensional plane to three dimensional space.
 

Ptolemy's Theorem

Ptolemy theorem is basically a relation between the four sides and the two diagonals of a cyclic quadrilateral. A cyclic quadrilateral is the one whose vertices exist on a common circle. 

A cyclic quadrilateral with vertices as A, B, C and D

Here, the bars denote the lengths of the line segments.

Let us discuss the proof of the theorem in detail:

Proof: Consider the cyclic quadrilateral ABCD and extend CD to P in such a way that ∠BAD = ∠CAP.

The quadrilateral ABCD is cyclic, so m ∠ABC + m ∠ADC = 180°. But, we know that ∠ADP forms a linear pair with ∠ADC. Hence, this implies that ∠ADP = ∠ABC.

This gives that ? ABC~ ? ADP (by AA similarity).

Now, AB/AD = BC/DP which means that DP = (AD) (BC)/ (AB).

Since the two angles subtend the same arc, so we have ∠ABD = ∠ACD. Moreover, ∠BAC + ∠CAD = ∠DAP + ∠CAD.

By cancelling the common angle on both sides we get,

∠BAD = ∠CAP.

Hence, ? BAD ~ ? CAP.

This gives AB/AC = BD/CP. Hence, CP = (AC) (BD)/ (AB).

But, CP = CD + DP. Making the required substitutions, we get

(AC) (BD)/ (AB) = CD + (AD)(BC)/ (AB).

Multiplying by AB on both sides gives,
(AC)(BD) = (AB)(CD) + (AD)(BC).

It may also be stated in another way. It says that the product of the lengths of the diagonals of a converse quadrilateral inscribed in a circle is equal to the sum of the products of lengths of the two pairs of its opposite sides, i.e.,

        |z₁-z₃| |z₂-z₄| = |z₁-z₂| |z₃-z₄|+|z₁-z₄| |z₂-z₃|

Ptolemy’s Inequality: Another result that follows from the Ptolemy’s theorem is the Ptolemy’s inequality. If A, B, C and D are any four points which are not necessarily concyclic then,

AB·CD + BC·AD ≥ AC·BD 

askIITians offers comprehensive study material which contains all the important topics of IIT JEE Mathematics. Questions like how to find out reflection of point with respect to line or inverse points of a circle or reflections of complex numbers have been aptly answered in the material. It is important to have thorough knowledge of this topic in order to remain competitive in the JEE.
 

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