Q3:

A tangent $PQ$ at a point $P$ of a circle of radius $5$ cm meets a line through the centre $O$ at a point $Q$ so that $OQ = 12$ cm. Length $PQ$ is :

Given:

• A circle with center $O$ and radius $OP = 5$ cm.
• A tangent $PQ$ at point $P$ on the circle.
• A line segment $OQ$ passing through the center $O$ such that $OQ = 12$ cm.

To find:

The length of the tangent $PQ$.

Step 1: Identifying the geometric relationship

According to the Theorem 10.1 of circles: "The tangent at any point of a circle is perpendicular to the radius through the point of contact."

Therefore, $OP \perp PQ$. This implies that $\angle OPQ = 90^\circ$.

Step 2: Applying the Pythagorean Theorem

Since $\triangle OPQ$ is a right-angled triangle with the right angle at $P$, we can apply the Pythagorean Theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In $\triangle OPQ$:

$OQ^2 = OP^2 + PQ^2$

Step 3: Substituting the known values

Given $OP = 5$ cm and $OQ = 12$ cm, we substitute these into the equation:

$(12)^2 = (5)^2 + PQ^2$

$144 = 25 + PQ^2$

Step 4: Solving for $PQ$

Subtract $25$ from both sides of the equation:

$PQ^2 = 144 - 25$

$PQ^2 = 119$

Taking the square root of both sides:

$PQ = \sqrt{119}$ cm

Final Answer: The length of the tangent $PQ$ is $\sqrt{119}$ cm.