Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Construction Procedure:
For the given circle, the tangent can be drawn as follows.
1. Draw a circle of 4 cm radius with centre “O”.
2. Again, take O as centre draw a circle of radius 6 cm.
3. Locate a point P on this circle
4. Join the points O and P through lines such that it becomes OP.
5. Draw the perpendicular bisector to the line OP
6. Let M be the mid-point of PO.
7. Draw a circle with M as its centre and MO as its radius
8. The circle drawn with the radius OM, intersect the given circle at the points Q and R.
9. Join PQ and PR.
10. PQ and PR are the required tangents.
It can be calculated manually as follows
In ∆PQO,
Since PQ is a tangent,
∠PQO = 90°. PO = 6cm and QO = 4 cm
Applying Pythagoras theorem in ∆PQO, 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
Therefore, the tangent length PQ = 4.47
To prove this, join OQ and OR represented in dotted lines.
From the construction,
∠PQO is an angle in the semi-circle.
We know that angle in a semi-circle is a right angle, so it becomes,
∴ ∠PQO = 90°
Such that
⇒ OQ ⊥ PQ
Since OQ is the radius of the circle with radius 4 cm, PQ must be a tangent of the circle. Similarly, we can prove that PR is a tangent of the circle.