Q1:

In each of the following, give also the justification of the construction: 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.

Given: A circle with center $O$ and radius $r = 6\text{ cm}$. A point $P$ located at a distance $OP = 10\text{ cm}$ from the center $O$.

To Find: Construct a pair of tangents from point $P$ to the circle and measure their lengths.

Step 1: Construction Procedure

1. Draw a circle with center $O$ and radius $6\text{ cm}$.

2. Mark a point $P$ such that $OP = 10\text{ cm}$.

3. Draw the line segment $OP$. Find the midpoint $M$ of $OP$ by drawing the perpendicular bisector of $OP$.

4. With $M$ as the center and $MO$ as the radius, draw a circle. This circle will intersect the original circle at two points, $T_1$ and $T_2$.

5. Join $PT_1$ and $PT_2$. These are the required tangents.

Step 2: Justification

To justify that $PT_1$ and $PT_2$ are tangents, we join $OT_1$.

In $\triangle OT_1P$, $\angle OT_1P$ is an angle in a semicircle. [By Thales' Theorem, the angle subtended by a diameter at the circumference is $90^\circ$].

Therefore, $\angle OT_1P = 90^\circ$.

Since $OT_1$ is a radius of the circle, $PT_1$ must be a tangent to the circle at $T_1$. [Since a line perpendicular to the radius at the point of contact is a tangent].

Step 3: Calculation of Tangent Length

In the right-angled triangle $\triangle OT_1P$:

$OP^2 = OT_1^2 + PT_1^2$ [Using Pythagoras Theorem]

Given $OP = 10\text{ cm}$ and $OT_1 = 6\text{ cm}$ (radius).

$10^2 = 6^2 + PT_1^2$

$100 = 36 + PT_1^2$

$PT_1^2 = 100 - 36$

$PT_1^2 = 64$

$PT_1 = \sqrt{64} = 8\text{ cm}$.

Final Answer: The length of each tangent is 8 cm.