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प्रश्न
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Give the justification of the construction.
उत्तर १
A pair of tangents to the given circle can be constructed as follows.
Step 1
Taking any point O of the given plane as centre, draw a circle of 6 cm radius. Locate a point P, 10 cm away from O. Join OP.
Step 2
Bisect OP. Let M be the mid-point of PO.
Step 3
Taking M as centre and MO as radius, draw a circle.
Step 4
Let this circle intersect the previous circle at point Q and R.
Step 5
Join PQ and PR. PQ and PR are the required tangents.
The lengths of tangents PQ and PR are 8 cm each.
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 6 cm). For this, 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.
उत्तर २
Given that
Construct a circle of radius 6 cm, and let a point P = 10 cm form its centre, construct the pair of tangents to the circle.
Find the length of tangents.
We follow the following steps to construct the given
Step of construction
Step: I- First of all we draw a circle of radius AB = 6 cm.
Step: II- Make a point P at a distance of OP = 10 cm, and join OP.
Step: III -Draw a right bisector of OP, intersecting OP at Q .
Step: IV- Taking Q as centre and radius OQ = PQ, draw a circle to intersect the given circle at T and T’.
Step: V- Joins PT and PT’ to obtain the require tangents.
Thus, PT and P'T' are the required tangents.
Find the length of tangents.
As we know that OT ⊥ PT and ΔOPT is right triangle.
Therefore,
OT = 6cm and PO = 10cm
In ΔOPT,
PT2 = OP2 - OT2
PT2 = 102 - 62
PT2 = 100 - 36
PT2 = 64
`PT = sqrt64=8`
Thus, the length of tangents = 8 cm.
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