Tangents on the given circle can be drawn as follow.
Step 1
Draw a circle of 4 cm radius with centre as O on the given plane.
Step 2 Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle a join OP.
Step 3
Bisect OP. Let M be the mid-point of PO.
Step 4
Taking M as its centre and MO as Its radius, draw a circle. Let it intersect the given circle at the points Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4.47 cm each.
In ΔPQO,
since PQ is a tangent,
∠PQO = 90°
PO = 6 cm
QO = 4 CM
Applying Pythagoras theorem in ΔPQO, we obtain
PQ² + QO² = PQ² ⇒ PQ² + (4)² = (6)² ⇒ PQ² + 16 = 36
PQ² = 36 – 16 ⇒ PQ² = 20 ⇒ PQ = 2√5
PQ = 4.47 cm
Justification
The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ and OR.
∠PQO is an angle in the semi- circle. We know that angle in a semi-circle is a right angle.
∴ ∠PQO = 90° ⇒ OQ ⊥PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.