Question

Tangents to the circle with centre 0, at the points A and B intersect at P, a circle
drawn with centre P passing through A, prove that the tangent at A to the circle with
centre P passes through O.​

Answer 1

Given :-

• Tangents to the circle with centre O, at the points A and B intersect at P,
• A circle drawn with centre P passing through A,

To Proved :-

• prove that the tangent at A to the circle with centre P passes through O.

Solution :-

given that, PA is a tangent to the circle with centre O.

Now, we have , In circle with centro O :-

→ AP = radius .

→ AP is a tangent to the radius.

So,

→ ∠PAO = 90° { Tangent is perpendicular to the radius. }

Than, in circle with centre P,

→ ∠OAP = 90° {Tangent (line l) is perpendicular to the radius AP .}

So, OA is perpendicular to the radius PA.

Therefore, OA is the tangent to the circle with centre P at A.

Hence, line segment OA coincide with line l.

∴ The tangent at A to the circle with centre P passes through O.

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Answer 2

Step-by-step explanation:

Diagram

Solution

GIVEN: Circles with centre O & O' , AC is tangent to the circle with centre o' at A. OO' meets AB at E.

TO PROVE THAT:

∠BAO=∠CAO

PROOF: In △O′AO

O'A = O'O ( being radii of the same circle with centre O')

So, ∠O′AO=∠O′OA = y……..(1)

O'A is perpendicular to tangent AC ( as theorem states that tangent to any circle at any point is perpendicular to the radius through that point.)

So, ∠O′AC= 90°…………..(2)

Eq(2) - Eq(1)

∠O′AC—∠O′A= ∠OAC = 90°— y ………..(3)

Now since ,

O is equidistant from A& B ( being radii of the same circle with centre O)

That implies that O lies on the perpendicular bisector of the segment AB.

Similarly, O' is equidistant from A & B( being radii of the same circle with centre O')

That implies that O' lies on the perpendicular bisector of the segment AB.

Hence conclude that OO' is perpendicular bisector of AB , at E ( by theorem )

Now, in

△AEO∠AEO = 90° ( proved above)

∠O′OA=y

( by eq (1) )

Therefore, third

∠OAE=90°—y………(4)

By comparing eq(3) & (4)

We have,

∠OAC=∠OAE

[Hence proved]