# 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.

Dear Student,

Tangents on the given circle can be drawn as follows.

**Step 1**

Draw a circle of 3 cm radius with the centre as O on the given plane.

**Step 2 **

Draw a circle of 6.5 cm radius taking O as its center. Locate a point P on this circle and join OP.

**Step 3**

Bisect OP. Let M be the mid-point of PO.

**Step 4**

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.

**Step 5 **

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.

In ΔPQO,

Since PQ is a tangent,

∠PQO = 90°

PO = 6.5 cm

QO = 3 cm

Applying Pythagoras theorem in ΔPQO, we obtain

PQ^{2} + QO^{2} = PQ^{2}

PQ^{2} + (3)^{2} = (6.5)^{2}

PQ^{2 }+ 9= 42.25

PQ^{2 }= 42.25 − 9

PQ^{2 }= 33.25

PQ =5.76 = 5.8

PQ = 5.8 cm

**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 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.

Regards

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