To construct a circle from given conditions
1. Let us draw a circle with radius = 6 cm with centre O.
2. Position a point P, which is 10 cm away from O.
3. Join the points O and P through line
4. Extend the perpendicular bisector of the line OP.
5. Let M be the mid-point of the line PO.
6. Take M as a centre and measure the length of MO
7. The length MO is taken as radius and draw a circle.
8. The circle drawn with the radius of MO, intersect the previous circle at point Q and R.
9. Join PQ and PR.
10. Therefore, PQ and PR are the required tangents.
We have to prove that PQ and PR are the tangents to the circle of radius 6cm 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 6 cm, PQ must be a tangent of the circle.
Similarly, we can prove that PR is a tangent of the circle.