Construct a tangent to a circle of radius 3 cm away from a point on the concentric circle of radius 6.5 cm and measure its length.
Tangents on the given circle can be drawn as follows.
Draw a circle of 3 cm radius with the centre as O on the given plane.
Draw a circle of 6.5 cm radius taking O as its center. Locate a point P on this circle and join OP.
Bisect OP. Let M be the mid-point of PO.
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.
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 5.8 cm each.
Since PQ is a tangent,
∠PQO = 90°
PO = 6.5 cm
QO = 3 cm
Applying Pythagoras theorem in ΔPQO, we obtain
PQ2 + QO2 = PQ2
PQ2 + (3)2 = (6.5)2
PQ2 + 9= 42.25
PQ2 = 42.25 − 9
PQ2 = 33.25
PQ =5.76 = 5.8
PQ = 5.8 cm
The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 3 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.