# Important Questions for Chapter 10 - Circles

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**Important questions based on NCERT syllabus for Chapter 10 - Circles:**

*Question 1*: In Fig., if AB = AC, prove that BE = EC

*Solution*:

Since tangents from an exterior point to a circle are equal in length

Therefore, AD = AF ... (i) [Tangents from A]

BD = BE ....(ii) [Tangents from B]

CE = CF ....(iii) [Tangents from C]

Now, AB = AC

⇒ AB – AD = AC – AD

[Subtracting AD from both sides]

⇒ AB – AD = AC – AF [Using (i)]

⇒ BD = CF

⇒ BE = CF [Using (ii)]

⇒ BE = CE [Using (iii)]

*Question 2*: PA and PB are tangents from P to the circle with center O. At point M which lies on circle, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN.

*Solution*:

We know that the tangents drawn from an external point to a circle are equal in length.

KA = KM .... (i) [From K]

and, NB = NM .... (ii) [From N]

Adding equations (i) and (ii), we get

KA + NB = KM + NM

⇒ AK + BN = KM + MN

⇒ AK + BN = KN

*Question 3*: Prove that the tangents at the extremities of any chord make equal angles with the chord.

*Solution*:

Let AB be a chord of a circle with center O, and let AP and BP be the tangents at A and B respectively.

Suppose the tangents meet at P .

Join OP. Suppose OP meets AB at C . We have to prove that ∠PAC = ∠PBC .

In triangles PCA and PCB , We have

PA = PB

∠APC = ∠BPC

[The tangents are equally inclined to line joining extrernal point and centre of circle.]

PC = PC [Common]

So, by SAS - criterion of congruence, we have

ΔPAC ≅ ΔPBC

⇒ ∠PAC = ∠PBC