To construct a circle from given conditions
1. Let us draw a circle of 4 cm radius with centre “O”.
2. Considering O as the centre draw a circle of radius 6 cm.
3. Position a point P on this circle
4. Join the points O and P through lines such that it becomes OP.
5. Extend 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.
From the construction, it is observed that PQ and PR are of length 4.47 cm each.
It can be calculated manually as follows
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
Hence, the tangent length PQ = 4.47
We have to prove that PQ and PR are the tangents to the circle of radius 4 cm with centre O.
Let us 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°
⇒ OQ ⊥ PQ
Since OQ is the radius of the circle with a radius of 4 cm, PQ must be a tangent of the circle.
Similarly, we can prove that PR is a tangent of the circle.