Following steps will be followed for constructing the tangents on the given circles:
1. Draw a line segment AB of 8 cm. Taking A and B as centre draw two circles of 4 cm and 3 cm radius. 
2. Bisect the line AB. Let mid-point of AB is C. Taking C as centre draw a circle of radius AC which will intersect the circles at point P, Q, R and S. Join BP, BQ, AS and AR. These are required tangents.

Here, AB = 5 cm, BC = 12 cm.
Let the radius of circle be x cm.

\(AC=\sqrt { (12)^{ 2 }+(5)^{ 2 } }\) 
\( =\sqrt { 144+25 } \)
\( =\sqrt { 169=13\quad } cm\\ \)
AC = AR + RC
AC = (5 - x) + (12 - x)
13 = 5 - x + 12 - x
2x = 17-13 = 4

Steps of construction :
1. Draw a line BC= 7·5 cm

2. Draw a perpendicular bisector of BC which cut the line BCat O.
3. Cut the line OA = 3·5 cm.
4. Join A to Band C
5. ABC is the given triangle.
6. Draw a ray BXmaking an acute angle.
7. Locate points B₁, B₂, BCon line segment BX.
8. Join B,C Draw a parallel line through B₄to B₇C intersecting extended line segment B Cat C.
9. Through C draw a line parallel to AC intersecting extended line segment AB at A'.
10.Hence, A'BC' is a required triangle.

12 cm

4 cm

4.1 cm

3.3 cm, 4.4 cm

Yes, the new triangle is also equilateral.

Given: A circle of radius 3.5 cm and a point P, 6.2 cm away from its centre. 
Required: A pair of tangents. 
Steps of Construction :

1. Draw a circle C (O, r) with centre O and radius 3.5 cm.
2. TakeapointP, such that OP = 6.2 cm.
3. Draw AB, the perpendicular bisector of OP and let it intersect OP in M. 
4. With M as centre and PM or OM as radius, draw another circle intersecting the given circle in T and T^'. 
5. Join PT and PT^'. 
6. Thus, PT and PT^' are the required tangents from point P to the circle C (O, r).

Steps of Construction:
(i) With O as centre a circle of radius 3 cm is drawn.
(ii) With same centre O another circle of radius 5 cm is drawn.
(iii) A point P is taken on outer circle and OP is joined.
(iv) Perpendicular bisector of OP is drawn intersecting OP at Q.
(v) With Q as centre and OQ as radius a circle is drawn intersecting the smaller circle at A and B.
(vi)  PA and PB is joined.
(vii) PA and PB are the required tangents.
 Length of tangent = 4 cm.

(i) Draw a line segment BC = 5.5 cm.
(ii) Perpendicular bisector XY of BC is drawn intersecting BC at L.
(iii) Point A is marked on XL such that AL=3 cm
(iv) AB and AC are joined.
(v) An acute ㄥCBP is drawn below BC.
(vi) On BP, points B₁ ,B₂ ,B₃ and B₄ are marked such that BB₁=B₁B₂=B₂B₃=B₃B₄
(vii) B₄C is joined
(viii) B₃C' is drawn parallel to B₄C interesting BC at C'
(ix) C'A' is drawn parallel to CA intersecting BA at A'