Q3:

In each of the following, give also the justification of the construction: Draw a circle of radius 3 cm. Take two points $P$ and $Q$ on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points $P$ and $Q$.

Given: A circle with center $O$ and radius $r = 3\text{ cm}$. A diameter is drawn and extended on both sides. Two points $P$ and $Q$ are marked on this extended diameter such that $OP = 7\text{ cm}$ and $OQ = 7\text{ cm}$.

To Find: Construct tangents from points $P$ and $Q$ to the circle and provide the geometric justification for the construction.

Steps of Construction:

Step 1: Draw a circle with center $O$ and radius $3\text{ cm}$.

Step 2: Draw a line passing through $O$ and mark points $P$ and $Q$ on this line such that $OP = 7\text{ cm}$ and $OQ = 7\text{ cm}$.

Step 3: Find the midpoint of $OP$. Let this be $M_1$. With $M_1$ as center and $M_1P$ as radius, draw a circle. This circle intersects the original circle at points $A$ and $B$.

Step 4: Join $PA$ and $PB$. These are the required tangents from $P$.

Step 5: Find the midpoint of $OQ$. Let this be $M_2$. With $M_2$ as center and $M_2Q$ as radius, draw a circle. This circle intersects the original circle at points $C$ and $D$.

Step 6: Join $QC$ and $QD$. These are the required tangents from $Q$.

Justification:

To justify the construction, consider the tangent $PA$. Join $OA$.

In $\triangle OAP$, the angle $\angle OAP$ is an angle in a semicircle (since the circle with diameter $OP$ passes through $A$).

[By Thales' Theorem/Angle in a semicircle property: An angle inscribed in a semicircle is a right angle.]

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

Since $OA$ is the radius of the original circle and $PA$ is a line segment perpendicular to the radius at its point of contact $A$, $PA$ must be a tangent to the circle.

[Theorem: A line drawn perpendicular to the radius at the point of contact is a tangent to the circle.]

The same logic applies to $PB$, $QC$, and $QD$.

Calculation of Tangent Length:

In right-angled triangle $\triangle OAP$:

$OP^2 = OA^2 + AP^2$ [Using Pythagoras Theorem]

$7^2 = 3^2 + AP^2$

$49 = 9 + AP^2$

$AP^2 = 40$

$AP = \sqrt{40} = 2\sqrt{10} \approx 6.32\text{ cm}$.

Since $OQ = OP = 7\text{ cm}$ and the radius $OA = OC = 3\text{ cm}$, by symmetry, the lengths of all tangents $PA, PB, QC, QD$ are equal to $\sqrt{40}\text{ cm}$.

Final Answer: The tangents $PA, PB, QC,$ and $QD$ have been constructed, each with a length of $\sqrt{40} \text{ cm} \approx 6.32 \text{ cm}$.