Ans. Given: A circle whose centre is O and radius is 6 cm and a point P is 10 cm away from its centre. 

To construct: To construct the pair of tangents to the circle and measure their lengths.

Steps of Construction:

(a) Join PO and bisect it. Let M be the mid-point of PO.
(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.
(c) Join PQ and PR.

Then PQ and PR are the required two tangents.
By measurement, PQ = PR = 8 cm

Justification: Join OQ and OR.

Since OQP and ORP are the angles in semicircles.

OQP  = ORP

Also, since OQ, OR are radii of the circle, PQ and PR will be the tangents to the circle at Q and R respectively.

 We may see that the circle with OP as diameter increases the given circle in two points. Therefore, only two tangents can be draw.

Ans. To construct: To construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its lengths. Also to verify the measurements by actual calculation.
 

Steps of Construction:
(a) Join PO and bisect it. Let M be the mid-point of PO.
(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the point Q and R.
(c) Join PQ.

Then PQ is the required tangent.

By measurement, PQ = 4.5 cm
By actual calculation,

PQ = 

=  = 

=  = 4.47 cm = 4.5 cm

Justification: Join OQ. Then PQO is an angle in the semicircle and therefore,

PQO = 

PQ  OQ

Since, OQ is a radius of the given circle, PQ has to be a tangent to the circle.

Ans. To construct: A circle of radius 3 cm and take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre and then draw tangents to the circle from these two points P and Q. 

Steps of Construction:
(a) Bisect PO. Let M be the mid-point of PO.
(b) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points A and B.
(c) Join PA and PB. Then PA and PB are the required two tangents.
(d) Bisect QO. Let N be the mid-point of QO.
(e) Taking N as centre and NO as radius, draw a circle. Let it intersects the given circle at the points C and D.
(f) Join QC and QD.

Then QC and QD are the required two tangents.

Justification: Join OA and OB.

Then PAO is an angle in the semicircle and therefore PAO = .

PA  OA

Since OA is a radius of the given circle, PA has to be a tangent to the circle. Similarly, PB is also a tangent to the circle.

Again join OC and OD.

Then QCO is an angle in the semicircle and therefore QCO = .

Since OC is a radius of the given circle, QC has to be a tangent to the circle. Similarly, QD is also a tangent to the circle.

Ans. To construct: A pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of  

Steps of Construction:

(a) Draw a circle of radius 5 cm and with centre as O.
(b) Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A with the help of compass.
(c) Draw a radius OB, making an angle of 120° (180° − 60°) with OA.
(d) Draw a perpendicular to OB at point B with the help of compass. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.

Justification: The construction can be justified by proving that ∠APB = 60°
By our construction
∠OAP = 90°
∠OBP = 90°
And ∠AOB = 120°
We know that the sum of all interior angles of a quadrilateral = 360°
∠OAP + ∠AOB + ∠OBP + ∠APB = 360°
90° + 120° + 90° + ∠APB = 360°
∠APB = 60°

Ans. To construct: A line segment of length 8 cm and taking A as centre, to draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Also, to construct tangents to each circle from the centre to the other circle. 

Steps of Construction:

(a) Bisect BA. Let M be the mid-point of BA.
(b) Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points P and Q.
(c) Join BP and BQ. Then, BP and BQ are the required two tangents from B to the circle with centre A.
(d) Again, Let M be the mid-point of AB.
(e) Taking M as centre and MB as radius, draw a circle. Let it intersects the given circle at the points R and S.
(f) Join AR and AS.

Then, AR and AS are the required two tangents from A to the circle with centre B.

Justification: Join BP and BQ.

Then APB being an angle in the semicircle is 

BP  AP

Since AP is a radius of the circle with centre A, BP has to be a tangent to a circle with centre A. Similarly, BQ is also a tangent to the circle with centre A.

Again join AR and AS.

Then ARB being an angle in the semicircle is 

AR  BR

Since BR is a radius of the circle with centre B, AR has to be a tangent to a circle with centre B. Similarly, AS is also a tangent to the circle with centre B.

Ans. To construct: A right triangle ABC with AB = 6 cm, BC = 8 cm and B =  BD is the perpendicular from B on AC and the tangents from A to this circle. 

Steps of Construction:
(a) Draw a right triangle ABC with AB = 6 cm, BC = 8 cm and B =  Also, draw perpendicular BD on AC.
(b) Join AO and bisect it at M (here O is the centre of circle through B, C, D).
(c) Taking M as centre and MA as radius, draw a circle. Let it intersects the given circle at the points B and E.
(d) Join AB and AE.

Then AB and AE are the required two tangents.

Justification: Join OE.

Then, AEO is an angle in the semicircle.

AEO = 

 AE  OE

Since OE is a radius of the given circle, AE has to be a tangent to the circle. Similarly, AB is also a tangent to the circle.

Ans. To construct: A circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle. 

Steps of Construction:
(a) Draw a circle with the help of a bangle.
(b) Take two non-parallel chords AB and CD of this circle.
(c) Draw the perpendicular bisectors of AB and CD. Let these intersect at O. Then O is the centre of the circle draw.
(d) Take a point P outside the circle.
(e) Join PO and bisect it. Let M be the mid-point of PO.
(f) Taking M as centre and MO as radius, draw a circle. Let it intersects the given circle at the points Q and R.
(g) Join PQ and PR.

Then PQ and PR are the required two tangents.

Justification: Join OQ and OR.

Then, PQO is an angle in the semicircle.

PQO = 

 PQ  OQ

Since OQ is a radius of the given circle, PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.