(i) The angle between a tangent to a circle and the radius drawn at the point of contact is

Justification: By theorem, the tangent at any point of a circle is perpendicular to the radius through the point of contact.

(ii) From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is

Justification: Using Pythagoras theorem:
\( r = \sqrt{OQ^2 – PQ^2} = \sqrt{25^2 – 24^2} = \sqrt{625 – 576} = \sqrt{49} = 7 \, \text{cm} \)

(iii) If AP and AQ are the two tangents to a circle with centre O so that \(\angle POQ = 110^\circ\), then \(\angle PAQ\) is equal to

Justification: In quadrilateral APOQ, \(\angle PAQ = 180^\circ – \angle POQ = 180^\circ – 110^\circ = 70^\circ\) (since \(\angle OAP = \angle OQP = 90^\circ\))

(iv) If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then \(\angle POA\) is equal to

Justification: \(\angle APB = 80^\circ\), so \(\angle APO = 40^\circ\). In right triangle OAP, \(\angle POA = 90^\circ – 40^\circ = 50^\circ\)

(v) In the figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B then \(\angle AOB\) =

Justification: OA bisects \(\angle COY\) and OB bisects \(\angle COY’\). Since XY ∥ X’Y’, \(\angle COY + \angle COY’ = 180^\circ\), so \(\angle AOB = 90^\circ\)

Solution:

Let O be the common center. AB is chord of larger circle touching smaller circle at P.

OP ⊥ AB (radius perpendicular to tangent at point of contact)

In right triangle OPA:

\( OA^2 = OP^2 + AP^2 \)

\( 5^2 = 3^2 + AP^2 \) ⇒ \( AP^2 = 25 – 9 = 16 \) ⇒ \( AP = 4 \, \text{cm} \)

AB = 2 × AP = 8 cm (since OP bisects the chord AB)

Solution:

Let ABCD be a parallelogram circumscribing a circle.

For a quadrilateral to circumscribe a circle, the sums of lengths of opposite sides must be equal.

Thus, AB + CD = AD + BC

But in parallelogram, AB = CD and AD = BC (opposite sides equal)

Therefore, 2AB = 2AD ⇒ AB = AD

Since all adjacent sides are equal, ABCD is a rhombus.

Solution:

Let the circle touch AB at E and AC at F.

From same external point, tangent lengths are equal:

AE = AF = x, BE = BD = 9 cm, CF = CD = 3 cm

Perimeter p = AB + BC + CA = (x+9) + 12 + (x+3) = 2x + 24

Semi-perimeter s = x + 12

Area = r × s = 3(x + 12)

Also by Heron’s formula: \( \sqrt{(x+12)(x)(3)(9)} = 3(x+12) \)

Squaring both sides: \( 27x(x+12) = 9(x+12)^2 \) ⇒ \( 3x = x+12 \) ⇒ \( x = 6 \, \text{cm} \)

Thus, AB = x + 9 = 15 cm, AC = x + 3 = 9 cm

Solution:

Construction Steps:

1. Draw circle with center O, radius 6cm
2. Mark point P 10cm from O
3. Draw perpendicular bisector of OP to find midpoint M
4. With M as center and MO as radius, draw circle intersecting first circle at Q and R
5. Join PQ and PR – these are the required tangents

Verification:

Length of tangent = \( \sqrt{OP^2 – r^2} = \sqrt{10^2 – 6^2} = \sqrt{64} = 8 \, \text{cm} \)

Measured lengths should match this calculation.

Solution:

Construction Steps:

1. Draw circle C1 with center O, radius 4cm
2. Draw concentric circle C2 with radius 6cm
3. Mark point P on C2
4. Join OP and find its midpoint M
5. With M as center and MO as radius, draw circle intersecting C1 at Q
6. Join PQ – this is the required tangent

Verification:

Length of tangent = \( \sqrt{OP^2 – r^2} = \sqrt{6^2 – 4^2} = \sqrt{20} = 2\sqrt{5} \, \text{cm} \approx 4.47 \, \text{cm} \)

Solution:

Construction Steps:

1. Trace the bangle to draw a circle (unknown center)
2. Draw two chords and their perpendicular bisectors to find center O
3. Mark external point P
4. Join OP and find its midpoint M
5. With M as center and MO as radius, draw circle intersecting original circle at Q and R
6. Join PQ and PR – these are the required tangents

Conclusion: Both tangents from an external point to a circle are equal in length.

Solution:

Given: ∠ABC = 90°, AB is diameter ⇒ ∠APB = 90° (angle in semicircle)

Let tangent at P meet BC at Q

∠BPQ = ∠BAP (angles in alternate segment)

But ∠BAP = ∠BCA (both complementary to ∠ABC)

Thus, ∠BPQ = ∠BCA ⇒ PQ ∥ AC (corresponding angles equal)

In ΔABC, P is midpoint of AC (since PQ ∥ AC and passes through center)

Thus, Q must be midpoint of BC (by midpoint theorem)

Solution:

Number of tangents: Exactly two tangents can be drawn from an external point to a circle.

Construction Steps:

1. Join OR and find its midpoint M
2. With M as center and MO as radius, draw circle intersecting given circle at P and Q
3. Join RP and RQ – these are the two required tangents

Verification: Both RP and RQ will be equal in length and perpendicular to OP and OQ respectively.

Solution:

Given: OP = radius = 5 cm, OQ = 13 cm

Since PQ is tangent, OP ⊥ PQ (radius perpendicular to tangent at point of contact)

In right triangle OPQ:

\( OP^2 + PQ^2 = OQ^2 \)

\( 5^2 + PQ^2 = 13^2 \)

\( 25 + PQ^2 = 169 \)

\( PQ^2 = 144 \)

\( PQ = 12 \, \text{cm} \)

Solution:

Construction Steps:

1. Draw a circle with center O and any radius
2. Draw a line l outside the circle (not intersecting the circle)
3. Draw perpendicular from O to line l, meeting at point P
4. With OP as distance, draw line m parallel to l – this will be tangent (touches at one point)
5. Draw another line n parallel to l at distance less than OP – this will be secant (intersects at two points)

Solution:

Given: Distance from center (d) = 15 cm, Radius (r) = 9 cm

Length of tangent (l) from external point is given by:

\( l = \sqrt{d^2 – r^2} = \sqrt{15^2 – 9^2} = \sqrt{225 – 81} = \sqrt{144} = 12 \, \text{cm} \)

Solution:

Let AB be diameter of circle with center O.

Let PA be tangent at A and QB be tangent at B.

Property: Tangent is perpendicular to radius at point of contact.

Thus, PA ⊥ OA and QB ⊥ OB

But OA and OB lie on same line AB (diameter)

Therefore, PA ⊥ AB and QB ⊥ AB

If two lines are both perpendicular to the same line, they are parallel to each other.

Hence, PA ∥ QB