Q5:

In each of the following, give also the justification of the construction: Draw a line segment $AB$ of length 8 cm. Taking $A$ as centre, draw a circle of radius 4 cm and taking $B$ as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Given: A line segment $AB$ of length $8\text{ cm}$. Two circles are drawn: one with center $A$ and radius $r_1 = 4\text{ cm}$, and another with center $B$ and radius $r_2 = 3\text{ cm}$.

To Find/Construct: Construct tangents to the circle centered at $A$ from point $B$, and tangents to the circle centered at $B$ from point $A$. Provide the geometric justification for the construction.

Step 1: Construction Procedure

1. Draw a line segment $AB = 8\text{ cm}$.

2. Draw a circle with center $A$ and radius $4\text{ cm}$.

3. Draw a circle with center $B$ and radius $3\text{ cm}$.

4. To construct tangents from $B$ to circle $A$: Find the midpoint $M$ of $AB$ by drawing the perpendicular bisector of $AB$.

5. With $M$ as the center and $MA$ as the radius, draw a circle. This circle intersects circle $A$ at points $P$ and $Q$.

6. Join $BP$ and $BQ$. These are the required tangents from $B$ to circle $A$.

7. To construct tangents from $A$ to circle $B$: Using the same midpoint $M$, draw a circle with radius $MA$. This circle intersects circle $B$ at points $R$ and $S$.

8. Join $AR$ and $AS$. These are the required tangents from $A$ to circle $B$.

Step 2: Justification

To justify the construction, consider the tangents from $B$ to circle $A$ (points $P$ and $Q$):

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

Since $M$ is the midpoint of $AB$, $MA = MB = MP$ (radii of the circle with center $M$). Thus, $\angle APB = 90^\circ$.

Since $AP$ is a radius of circle $A$ and $\angle APB = 90^\circ$, it follows that $BP$ must be a tangent to circle $A$ at point $P$. [A line perpendicular to the radius at the point of contact is a tangent].

The same logic applies to $BQ$, $AR$, and $AS$.

Step 3: Verification of Tangent Lengths

In $\triangle APB$, by the Pythagorean theorem: $BP^2 = AB^2 - AP^2$.

$BP^2 = 8^2 - 4^2 = 64 - 16 = 48$.

$BP = \sqrt{48} = 4\sqrt{3} \approx 6.93\text{ cm}$.

In $\triangle ARB$, by the Pythagorean theorem: $AR^2 = AB^2 - BR^2$.

$AR^2 = 8^2 - 3^2 = 64 - 9 = 55$.

$AR = \sqrt{55} \approx 7.42\text{ cm}$.

Final Answer: The tangents have been constructed such that the angle between the radius and the tangent is $90^\circ$, confirming the geometric validity of the construction. The lengths of the tangents from $B$ to circle $A$ are $4\sqrt{3}\text{ cm}$ and from $A$ to circle $B$ are $\sqrt{55}\text{ cm}$.