Question

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, height PQ is :

• A. 12 cm
• B. 13 cm
• C. 8.5 cm
• D. \(\sqrt{119} \, \text{cm}\)

Hint:

Remember that the radius drawn to the point of tangency is always perpendicular to the tangent line. This creates a right-angled triangle, allowing the use of the Pythagorean theorem to solve for unknown lengths.

Answer 1

The Correct Option is D

Step 1: Visualize the geometry and identify key properties.
We have a circle with center O.
P is a point on the circle, so OP is the radius. The radius of the circle (OP) is given as 5 cm.
PQ is a tangent to the circle at point P.
A line passes through the center O and meets the tangent at point Q. The distance OQ is given as 12 cm.
Step 2: Apply the theorem related to tangents and radii.
A fundamental property of circles is that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore, the angle $\angle OPQ$ is a right angle ($90^\circ$).
This means that triangle $\triangle OPQ$ is a right-angled triangle, with OQ as the hypotenuse.
Step 3: Use the Pythagorean theorem to find the length of PQ.
In a right-angled triangle $\triangle OPQ$, according to the Pythagorean theorem:
$OP^2 + PQ^2 = OQ^2$ We are given:
$OP = 5$ cm
$OQ = 12$ cm
Substitute the known values into the equation:
$5^2 + PQ^2 = 12^2$
$25 + PQ^2 = 144$
Now, solve for $PQ^2$:
$PQ^2 = 144 - 25$
$PQ^2 = 119$
Finally, find PQ:
$PQ = \sqrt{119}$ cm
Step 4: Compare the result with the given options.
The calculated length of PQ is $\sqrt{119}$ cm, which matches option (4). (4) \( \sqrt{119} \text{ cm} \)