If a tangent to a circle from a point P meets the circle at A with AP = 15 cm. Given that the radius of the circle is 8 cm, find the distance of P from the centre of the circle.
17 cm
Finding Distance of Point P from Circle Center
The problem asks us to find the distance of an external point P from the center of a circle. We are given the length of the tangent segment from P to the circle and the radius of the circle.
Understanding the Geometry Problem
Let's visualize the scenario:
- We have a circle with a center, let's call it O.
- There is a point P outside the circle.
- A line segment from P touches the circle at exactly one point, say A. This segment AP is the tangent to the circle at A.
- We are given the length of this tangent segment, AP = 15 cm.
- We are also given the radius of the circle, OA = 8 cm.
- We need to find the distance of point P from the center O, which is the length of the segment OP.
Key Circle Property: Tangent and Radius
A crucial property in circle geometry is the relationship between a tangent and the radius at the point of contact. The tangent at any point of a circle is perpendicular to the radius through the point of contact.
In our case, the tangent is AP and the radius is OA. The point of contact is A. Therefore, the angle formed by the radius OA and the tangent AP at A is 90 degrees. So, $\angle \text{OAP} = 90^\circ$.
Applying Pythagoras Theorem to Find Distance
Since $\angle \text{OAP} = 90^\circ$, the triangle OAP is a right-angled triangle. The sides of this triangle are:
- OA (radius) = 8 cm
- AP (tangent length) = 15 cm
- OP (distance of P from center) - This is the hypotenuse as it is opposite the right angle.
According to the Pythagoras theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
For triangle OAP, the theorem states:
$\text{OP}^2 = \text{OA}^2 + \text{AP}^2$
Step-by-Step Calculation
Now, let's substitute the given values into the equation:
$\text{OP}^2 = (8 \text{ cm})^2 + (15 \text{ cm})^2$
Calculate the squares of the given lengths:
$(8 \text{ cm})^2 = 8 \times 8 \text{ cm}^2 = 64 \text{ cm}^2$
$(15 \text{ cm})^2 = 15 \times 15 \text{ cm}^2 = 225 \text{ cm}^2$
Now, add these values:
$\text{OP}^2 = 64 \text{ cm}^2 + 225 \text{ cm}^2$
$\text{OP}^2 = 289 \text{ cm}^2$
To find OP, we need to take the square root of 289:
$\text{OP} = \sqrt{289 \text{ cm}^2}$
The square root of 289 is 17.
$\text{OP} = 17 \text{ cm}$
Thus, the distance of point P from the center of the circle is 17 cm.
Comparing Result with Options
Let's check our calculated distance against the given options:
- Option 1: 20 cm (Incorrect)
- Option 2: 12 cm (Incorrect)
- Option 3: 17 cm (Correct)
- Option 4: 15 cm (Incorrect)
Our calculated distance, 17 cm, matches Option 3.
Revision Table: Circle Tangent Properties
| Concept | Description | Relevance to Problem |
|---|---|---|
| Tangent to a Circle | A line that touches the circle at exactly one point. | AP is the tangent segment. |
| Radius at Point of Contact | The radius drawn from the center to the point where the tangent touches the circle. | OA is the radius at the point of contact A. |
| Perpendicularity of Tangent and Radius | The tangent is always perpendicular ($\angle 90^\circ$) to the radius at the point of contact. | Forms a right-angled triangle OAP. |
| Pythagoras Theorem | In a right-angled triangle, $a^2 + b^2 = c^2$, where c is the hypotenuse. | Used to find the unknown side OP in $\triangle$OAP. |
Additional Information: Circle Geometry Concepts
Understanding tangents and radii is fundamental in circle geometry. Here are a few more related points:
- Length of Tangents from an External Point: If two tangents are drawn to a circle from an external point, their lengths from the external point to the points of contact are equal.
- Secant: A line that intersects a circle at two distinct points. A tangent can be thought of as a limiting case of a secant where the two intersection points coincide.
- Chord: A line segment connecting two points on the circle.
- Converse of Tangent-Radius Property: If a line passes through the end point of a radius on the circle and is perpendicular to it, then the line is a tangent to the circle.
These concepts are often used together to solve more complex geometry problems involving circles, tangents, and external points.
Important Questions from Circles, Chords and Tangents
In a circle with a radius of 10 cm. XY and PQ are two parallel chords 12 cm and 16 cm in length, respectively. The two chords are situated on the opposite sides of the centre. The distance between the chords is:
Find the equation of the tangents to the circle x2 + y2 = 9 at x = 2.
If a chord of length 24 cm is at a distance of 5 cm from centre, then find the radius of the circle.
Let \( C_1 \) and \( C_2 \) be two circles which do not externally touch and intersect each other and \( O_1 \), and \( O_2 \) be the centers of the circles, respectively. Let AB be the common transverse tangent to the circles such that P, Q are the points of tangency respectively to \( C_1 \), \( C_2 \). Let R be the point of intersection of \( O_1 O_2 \) and AB. If \( \angle PO_1R = 60^\circ \), find \( \angle QO_2R \) and \( \angle QRO_2 \) respectively.
Find the length of the direct common tangent to two circles with radii equal to $6$ cm and $18$ cm and the distance between their centers which is equal to $30$ cm.
