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CIRCLE

  • Sum based on Theorem – The lengths

of two tangents drawn from

an external point to a circle are equal.

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[The lengths of the two tangents from an external point to a circle are equal]

[The lengths of the two tangents from an external point to a circle are equal]

m

l

A

B

C

D

Q. Point A is a common point of contact of

two externally touching circles and line l

is a common tangent to both the circles

touching at B and C. Line m is another

common tangent at A and it intersects

BC at D. Prove : (i) ∠BAC = 90o

(ii) Point D is the midpoint of seg BC.

A

B

C

D

Proof.

In ΔBDA,

DB = DA

…(i)

∴

∠DBA

=

∠DAB

Let,

∠DBA

=

∠DAB

=

xº

…(ii)

x

In ΔDAC,

DC = DA

…(iii)

∴

∠DCA

=

∠DAC

D

Let,

∠DAC

=

∠DCA

=

yº

…(iv)

y

y

∠BAC is made up of

two angles

i.e. ∠BAD and ∠DAC

∠BAD belongs to ΔBAD

We know, length of two tangents drawn from an external point to a circle are equal

We know, angles opposite to equal sides are equal

Let us first focus on ∠BAD

Let us first focus on ∠DAC

∠DAC belongs to ΔDAC

We know, length of two tangents drawn from an external point to a circle are equal

We know, angles opposite to equal sides are equal

x

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m

l

A

B

C

D

Q. Point A is a common point of contact of

two externally touching circles and line l

is a common tangent to both the circles

touching at B and C. Line m is another

common tangent at A and it intersects

BC at D. Prove : (i) ∠BAC = 90o

(ii) Point D is the midpoint of seg BC.

∠BAC

=

∠DAB

+

∠DAC

[Angle Addition property]

∴

∠BAC

=

(x

…(v)

[From (ii) and (iv)]

+

Proof.

∠DBA = ∠DAB = x …(ii)

x

y

y

x

y)º

∠DCA = ∠DAC = y …(iv)

In ΔABC,

∠ABC

+

∠ACB

+

∠BAC

= 180º

[Angle sum property]

x

+

y

+

x

+

y

= 180

[From (ii), (iv), (v)]

∴

2x

+

2y

= 180

Consider ΔABC

∴

x + y

=

90

∴

∠BAC

=

90º

[From (v)]

Observe ∠BAC

∴

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m

l

A

B

C

D

Q. Point A is a common point of contact of

two externally touching circles and line l

is a common tangent to both the circles

touching at B and C. Line m is another

common tangent at A and it intersects

BC at D. Prove : (i) ∠BAC = 90o

(ii) Point D is the midpoint of seg BC.

Proof.

x

y

y

x

DB = DA

[from (i)]

DC = DA

[from (iii)]

∴

DB = DC

[from (i) & (iii)]

∴

D is the midpoint of seg BC.

DB = DA

…(i)

DC = DA

…(iii)