Line (Segment) from the Center to a Chord

Line(Segment) from the Center to a Chord

Let us investigate a line (line segment) drawn from the center of the circle to a chord. BC is a chord of the circle with center at O. D is a point on the chord. What do you notice about the lengths BD and CD when the angle is 90℃.
Move the point D on the chord and look for a situation when the angle is 90℃ at this point what can you say about the lengths BD and CD, are they equal? You will observe that BD = CD. Here we can make the following conjecture:

• The perpendicular from the center of a circle to a chord bisects the chord.

The converse of the above conjecture is also true:

• If a line (line segment) is drawn from the center of a circle to the midpoint of a chord, then the line is perpendicular to the chord.

Let us prove the conjecture 1 above. This proof is based on the congruency of the triangles.

       Given: OD ⊥ BC

       Construction: Join OB and OC.

       Strategy: If we can show that △BDO 
       and △CDO are congruent then the 
       sides BD and CD must be equal.

       To prove: BD = CD

       Proof:

          In △ BDO and △ CDO

          OB = OC (Radii of the circle)

          OD = OD (Common Sides)

         ∠BDO = ∠CDO (given , both 90)

      Hence, △ BDO ≅ △ CDO

      Therefor, BD = CD (Corresponding Parts of Congruent Triangles)

Circle Vocabulary (वृत्त शब्दावली)

Circle Vocabulary

Circle: The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle. The fixed point is called the center of the circle and the fixed distance is called the radius of the circle.
Chord: If you take two points A and B on the circle, then the line segment joining AB is called the chord of the circle.
Diameter: A chord that passes through the center of the circle is called the diameter of the circle. Diameter is the longest chord of the circle and it is twice the length of the radius of the circle.
Arc: A piece of the circle between two points is called an arc.
Circumference: The length of the complete circle is called its circumference.
Segment: The region between a chord and either of its arc is called a segment.
Sector: The region between an arc and the radii joining the endpoints of the arc to the center of the circle is called a sector.

All the above are shown in the applet below, you may move the points on the two circles and see the effects.

वृत्त शब्दावली

वृत्त : एक तल पर उन सभी बिन्दुओं का समूह , जो तल के एक स्थिर बिन्दु से एक स्थिर दूरी पर स्थित हों , एक वृत्त कहलाता है। स्थिर बिन्दु को वृत्त का केन्द्र कहते हैं तथा स्थित दूरी को वृत्त की त्रिज्या कहते हैं।
जीवा : यदि एक वृत्त पर दो बिन्दु A और B लें , तो रेखाखण्ड AB वृत्त की जीवा कहलाता है।
व्यास : वह जीवा जो वृत्त के केन्द्र से होकर जाती है , उसे वृत्त का व्यास कहते हैं। व्यास , वृत्त की सबसे बड़ी जीवा होती है। व्यास की लंबाई वृत्त की त्रिज्या से दो गुनी होती है।
चाप : दो बिन्दुओं के बीच के वृत्त के भाग को चाप कहते हैं।
परिधि : संपूर्ण वृत्त की लंबाई को उस वृत्त की परिधि कहते हैं।
वृत्तखण्ड : जीवा और प्रत्येक चाप के मध्य क्षेत्र को वृत्तखण्ड कहते हैं।
त्रिज्यखण्ड : केन्द्र को एक चाप के सिरों से मिलाने वाली त्रिज्याओं एवं चाप के बीच के क्षेत्र को त्रिज्यखण्ड कहते हैं।

नीचे दी गयी एपलेट में उपर परिभाषित सभी अवधारणाओं को दर्शाया गया है , दोनों वृत्त में दिखाए गए बिन्दुओं को हिला कर आप उनका प्रभाव देख सकते हैं।

Locate the Center of a given Circle

किसी दिए गए वृत्त के केन्द्र का पता लगाना

हमें एक वृत्त दिया गया है जिसके केन्द्र बिन्दु की जानकारी नहीं दी गयी है। हमें इस वृत्त के केन्द्र बिन्दु की स्थिति ज्ञात करनी है। यह कार्य निम्न चरणों में किया जा सकता है :

चरण 1 – किसी चूड़ी या कटोरी की मदद से एक वृत्त बनाएं। यह वृत्त हमें दिया गया है।

चरण 2 – दिए गए वृत्त की कोई दो असमान्तर जीवाएं PQ और LM खींचिए।

चरण 3 – PQ का लंबार्धक खींचें।

चरण 4 – LM का लंबार्धक खींचें।

चरण 5 – चरण 3 व चरण 4 के लंबार्धकों के प्रतिच्छेद बिन्दु O को अंकित करें।

चरण 6 – बिन्दु O दिए गए वृत्त का केन्द्र है।

नीचे एपलेट में बिन्दु P की मदद से वृत्त को छोटा – बड़ा किया जा सकता है।

To locate the center of a given circle

We are given a circle location of whose center is not known. We need to locate the center of this circle. This can be done in the following steps:

Step 1 – Draw a circle with the help of a bangle or a bowl.

Step 2 – Draw any two non – parallel chords PQ and LM of the given circle.

Step 3 – Construct the perpendicular bisector of the chord PQ.

Step 4 - Construct the perpendicular bisector of the chord LM.

Step 5 – Mark the point of intersection O of the two bisectors of step 3.

Step 6 – O is the center of the given circle.

In the applet below, use point P to change the size of the circle.

Nine Point Conics

Nine point circle of a triangle ABC is a circle passing through the mid points of sides of the triangle (G, H, I), feet of the perpendicular (D, E , F) drawn from the vertex to the opposite sides and the mid-points (K , L, M) of the distance from the orthocenter (N) to the three vertices of the triangle.

The concept of a nine point circle can be generalized to a nine point ellipse or a nine point hyperbola if we consider a general cevian instead of altitude. A cevian is any segment drawn from the vertex of a triangle to the opposite side. Cevians with special properties include altitudes, angle bisectors, and medians.

Consider three concurrent cevians with cevian point P, locate mid-points E, F and G of the segments from cevian point to the vertices of the triangle. Also locate the feet of the cevian K , N and L. If we draw a conic through any five of the above points, we will get an ellipse and it will also pass through the sixth point.

Now locate the mid points of the three sides of the triangle, we will find that these points also fall on the ellipse constructed above. The conic remains an ellipse when the feet of cevians lie on the sides of the triangle and converts to a nine point hyperbola when the feet of the cevians lie on the extensions of the sides.

Now locate centroid (G₁) of the triangle ABC and centre (N₁) of the conic , interestingly , the cevian point P , G₁ and N₁ lie on the same straight line with N₁P = 3 N₁G₁ . This is also the generalization of Euler Line.

Perpendicular from the Center to a Chord

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Vecten Point

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