Question

How many circle theorems are there ? What are they ?

Urgent , correct quality answer needed .​

Answer 1

Circles have different angle properties, described by theorems . There are Five circle theorems.

• The angle at the centre is twice the angle at the circumference.
• The angle in a semicircle is a right angle.
• Angles in the same segment are equal.
• Opposite angles in a cyclic quadrilateral sum to 180°
• The angle between the chord and the tangent is equal to the angle in the alternate segment.

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Answer 2

$\large\orange{\underline{{\boxed{\textbf{ \:Required\:Answer\: }}}}}$

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${\bold{\underline{\underline{1st \:Theorem \: :}}}}$

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given : A circle with centre O and a tangent XY to the circle at a point P.

[ Refer the 1st attachment for the diagram ]

To prove : Tangent is perpendicular to the radius at its end point.

i.e., OP ⊥ XY

Proof : To the contrary, let us assume that OP is not perpendicular to XY.

Let 'Q' be any point on XY.

Join OQ.

The point Q must lie outside the circle.

[∵ If it lies inside the circle, then XY becomes a secant and not a tangent]

∴ OQ is longer than OP.

[∵ OP is radius and OQ is the line segment joining an exterior point Q with the centre.]

∴ OQ > OP.

This must be true for all points on XY other than P.

∴ OP is the shortest of all distances of the centre 'O' to the points on XY.

So our assumption that OP is not perpendicular to XY is false.

∴ OP ⊥ XY

Hence Proved.

${\bold{\underline{\underline{2nd \:Theorem \: :}}}}$

The lengths of tangents drawn from an external point to a circle are equal.

Given : A circle with centre 'O', 'P' is exterior point. PA & PB are 2 tangents to the circles from P.

[ Refer the 2nd attachment for the diagram ]

To Prove : PA = PB

Proof : Join O, A, O, B and O, P.

∠OBP = ∠OAP = 90°

[angle between tangent and radius is 90°]

Now in ∆OAP, ∆OBP

OA = OB (radii of the same circle)

OP = OP (common)

∆OAP ≅ ∆OBP (R.H.S. congruence)

∴ PA = PB

[corresponding parts of congruent triangles are equal (C.P.C.T)]

Thus lengths of the tangents drawn from an external point to a point to a circle are equal. (Q.E.D)

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