Maths 1 Marks Question

Given: Two circles are touching each other externally

We know that if the circles touch each other externally, distance between their centres is equal to the sum of their radii.

⇒distance between their centres = 5.5 cm + 4.2 cm = 9.7 cm

Given: Two circles are touching each other internally.

∵The distance between the centres of the circles touching internally is equal to the difference of their radii.

⇒distance between their centres = 4.8 cm – 3.5 cm = 1.3 cm

Let BC and DE be the parallel tangents to a circle centered at A with point of contact O and H respectively. On joining OH, we find OH is the diameter of the circle.∠ BOA = 90° = ∠ DHA {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}
Distance between BC and DE = OH
∵ OH is perpendicular to BC and DE.
OH = 2 × 4.5 cm = 9 cm

In triangle MOR and triangle NOR,
MR = NR {∵Tangents from same external point are congruent to each other.}
OR = OR {Common}
OM = ON {Radius of the circle}
⇒ ΔMOR ≅ ΔNOR {By SSS}
⇒ ∠ROM = ∠RON
And ∠MRO = ∠NRO {C.P.C.T.}
Hence proved that seg OR bisects ∠MRNas well as ∠MON.

Similarly, ∠NRO = 30°
⇒∠MRN = ∠ MRO + ∠NRO = 30° + 30° = 60°

In triangle ABC right-angled at A,
AB = CA = 6 cm
⇒∠ABC = ∠ACB {Angles opposite to equal sides are equal}
⇒∠ABC + ∠ACB + ∠ BAC = 180° {Angle sum property of the triangle}
⇒ 2∠ABC = 90° {∵ ∠ BAC = 90°}
⇒ ∠ABC = 45°

CA is the radius of the circle which is perpendicular to the tangent AB.

So, the perpendicular distance of line AB from C = CA = 6 cm

ere CA is the radius of the circle and A is the point of contact of the tangent AB.

⇒ ∠CAB = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.

The perpendicular distance of O from P = radius of the circle = 9 cm.
(2)Q lies in the interior of the circle because P lieing on the circumference of the circle is at a distance of 9 cm.
(3)Position of R is not specified.

self

By theorem on interesecting chords,
PN ×PM=PR ×PS... (I)
let PN = x. ∴PM = 11 - x
substituting the values in (I),
x (11 - x) = 6 × 4
∴ 11x - x²- 24 = 0
∴ x² - 11x + 24 = 0
∴ (x - 3) (x - 8) = 0
∴ x - 3 = 0 or x - 8 = 0
∴ x = 3 or x = 8
∴ PN = 3 or PN = 8

PS²= PQ × PR .... tangent secant segments
theorem
= PQ × (PQ + QR)
= 3.6 × [3.6 + 6.4]
= 3.6 × 10
= 36
∴PS = 6

260°

8 cm

8 cm

Given: Two circles are touching each other externally
We know that if the circles touch each other externally, distance between their centres is equal to the sum of their radii.
⇒distance between their centres = 5.5 cm + 4.2 cm = 9.7 cm

Given: Two circles are touching each other internally.
∵The distance between the centres of the circles touching internally is equal to the difference of their radii.
⇒distance between their centres = 4.8 cm – 3.5 cm = 1.3 cm

Let BC and DE be the parallel tangents to a circle centered at A with point of contact O and H respectively. On joining OH, we find OH is the diameter of the circle.∠ BOA = 90° = ∠ DHA {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}
Distance between BC and DE = OH
∵ OH is perpendicular to BC and DE.
OH = 2 × 4.5 cm = 9 cm

In triangle MOR and triangle NOR,
MR = NR {∵Tangents from same external point are congruent to each other.}
OR = OR {Common}
OM = ON {Radius of the circle}
⇒ ΔMOR ≅ ΔNOR {By SSS}
⇒ ∠ROM = ∠RON
And ∠MRO = ∠NRO {C.P.C.T.}
Hence proved that seg OR bisects ∠MRNas well as ∠MON.

Similarly, ∠NRO = 30°
⇒∠MRN = ∠ MRO + ∠NRO = 30° + 30° = 60°

Here OM is the radius of the circle and M and N are the points of contact of MR and NR respectively.
⇒ ∠RMO = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.
In triangle ORM right-angled at M,
Given that OR = 10 cm and OM = 5 cm {Radius of the circle}
OR² = OM² + RM² {Using Pythagoras theorem}
⇒MR² = 10² -5²
⇒MR² = 100 - 25
⇒ MR = √75
⇒ MR = 5√3 cm
Also, RN = 5√3 cm {∵ Tangents from the same external point are congruent to each other.}

In triangle ABC right-angled at A,
AB = CA = 6 cm
⇒∠ABC = ∠ACB {Angles opposite to equal sides are equal}
⇒∠ABC + ∠ACB + ∠ BAC = 180° {Angle sum property of the triangle}
⇒ 2∠ABC = 90° {∵ ∠ BAC = 90°}
⇒ ∠ABC = 45°

In triangle ABC right-angled at A,

Given AB = 6 cm and CA = 6 cm
BC² = AB² + CA² {Using Pythagoras theorem}
⇒ BC² = 6² + 6²
⇒ BC² = 36 + 36
⇒ BC = √72
⇒ BC = 6√2 cm

CA is the radius of the circle which is perpendicular to the tangent AB.
So, the perpendicular distance of line AB from C = CA = 6 cm

ere CA is the radius of the circle and A is the point of contact of the tangent AB.
⇒ ∠CAB = 90° Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.

The perpendicular distance of O from P = radius of the circle = 9 cm.
(2)Q lies in the interior of the circle because P lieing on the circumference of the circle is at a distance of 9 cm.
(3)Position of R is not specified.

By theorem on interesecting chords,
PN ×PM=PR ×PS... (I)
let PN = x. ∴PM = 11 - x
substituting the values in (I),
x (11 - x) = 6  × 4
∴ 11x - x²- 24 = 0
∴ x² - 11x + 24 = 0
∴ (x - 3) (x - 8) = 0
∴ x - 3 = 0 or x - 8 = 0
∴ x = 3 or x = 8
∴ PN = 3 or PN = 8

PS²= PQ  × PR .... tangent secant segments
theorem
= PQ  × (PQ + QR)
= 3.6  × [3.6 + 6.4]
= 3.6  × 10
= 36
∴PS = 6

8 cm

8 cm

260°