Question

Construct a tangent to a circle of radius $4\mathrm{cm}$ from a point on the concentric circle of radius $6\mathrm{cm}$ and measure its length.

Also verify the measurement by actual calculation.

Give the justification of the construction.

Answer 1

Steps for construction:

$1$ Let us draw a circle of $4\mathrm{cm}$ radius with centre “$\mathrm{O}$”.

$2$ Considering $\mathrm{O}$ as the centre draw a circle of radius $6\mathrm{cm}$.

Position a point $\mathrm{P}$ on this circle.

$3$ Join the points $\mathrm{O}$ and $\mathrm{P}$ through a line such that it becomes $\mathrm{OP}$.

$4$ Extend the perpendicular bisector to the line $\mathrm{OP}$.

Let $\mathrm{M}$ be the mid-point of $\mathrm{OP}$.

$5$ Draw a circle with $\mathrm{M}$ as its centre and $\mathrm{MO}$ as its radius.

$6$ The circle drawn with the radius $\mathrm{MO}$, intersects the given circle at the points $\mathrm{Q}$ and $\mathrm{R}$.

$7$ Join $\mathrm{PQ}$ and $\mathrm{PR}$.

Hence $\mathrm{PQ}$ and $\mathrm{PR}$ are the required tangents.

From the construction, it is observed that $\mathrm{PQ}$ and $\mathrm{PR}$ are of length $4.47\mathrm{cm}$ each.

It can be calculated manually as follows:

In $∆\mathrm{PQO}$,

Since $\mathrm{PQ}$ is a tangent,

$\angle \mathrm{PQO}=90°$. $\mathrm{PO}=6\mathrm{cm}$ and $\mathrm{QO}=4\mathrm{cm}$.

Applying Pythagoras theorem in $∆\mathrm{PQO}$, we obtain ${\mathrm{PQ}}^2+{\mathrm{QO}}^2={\mathrm{PO}}^2$

${\mathrm{PQ}}^2+4^2=6^2$

${\mathrm{PQ}}^2+16=36$

${\mathrm{PQ}}^2=36-16$

${\mathrm{PQ}}^2=20$

$\mathrm{PQ}=2\sqrt{5}$

$\mathrm{PQ}=4.47\mathrm{cm}$

Similarly, $\mathrm{PR}=4.47\mathrm{cm}$

Hence, the tangents length $\mathrm{PQ}=4.47\mathrm{cm}$ and $\mathrm{PR}=4.47\mathrm{cm}$

Justification

We have to prove that $\mathrm{PQ}$ and $\mathrm{PR}$ are the tangents to the circle of radius $4\mathrm{cm}$ with centre $\mathrm{O}$.

Let us join $\mathrm{OQ}$ and $\mathrm{OR}$ represented in dotted lines.

From the construction,

$\angle \mathrm{PQO}$ is an angle in the semi-circle.

We know that angle in a semi-circle is a right angle, so it becomes,

$\angle \mathrm{PQO}=90°$

Such that

$\Rightarrow \mathrm{OQ}\perp \mathrm{PQ}$

Since $\mathrm{OQ}$ is the radius of the circle with a radius of $4\mathrm{cm}$, $\mathrm{PQ}$ must be a tangent of the circle.

Similarly, we can prove that $\mathrm{PR}$ is a tangent of the circle.