Understanding the Circle Geometry Problem

The question asks us to find the distance of a point P from the center O of a circle. We are given information about a tangent drawn from P that touches the circle at point Q. We know the length of the tangent segment PQ and the radius of the circle OQ.

Here's what we are given:

• O is the center of the circle.
• A tangent is drawn from point P.
• The tangent touches the circle at point Q.
• The length of the tangent segment PQ is 24 cm.
• The radius of the circle OQ is 10 cm.
• We need to find the value of OP.

Key Geometric Property: Radius and Tangent

A fundamental property in circle geometry states that the radius drawn to the point of tangency is perpendicular to the tangent line at that point. In this problem, OQ is the radius and PQ is the tangent segment at Q. Therefore, the radius OQ is perpendicular to the tangent PQ at point Q.

This means the angle formed at Q, \(\angle OQP\), is a right angle, i.e., \(\angle OQP = 90^\circ\).

Forming a Right-Angled Triangle

Since \(\angle OQP = 90^\circ\), the triangle \(\triangle OQP\) is a right-angled triangle. The sides of this triangle are:

• OQ (Radius) = 10 cm
• PQ (Tangent segment) = 24 cm
• OP (Distance from center to point P)

In the right-angled triangle \(\triangle OQP\), OP is the side opposite to the right angle Q. Therefore, OP is the hypotenuse of the triangle.

Applying the Pythagorean Theorem

The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

For \(\triangle OQP\), the theorem can be written as:

\[OP^2 = OQ^2 + PQ^2\]

Step-by-Step Calculation of OP

Now, we substitute the given values of OQ and PQ into the equation:

\[OP^2 = (10 \, \text{cm})^2 + (24 \, \text{cm})^2\]

Calculate the squares of OQ and PQ:

\[OP^2 = 100 \, \text{cm}^2 + 576 \, \text{cm}^2\]

Add the squared values:

\[OP^2 = 676 \, \text{cm}^2\]

To find OP, we need to take the square root of 676:

\[OP = \sqrt{676 \, \text{cm}^2}\]

We know that \(26 \times 26 = 676\). Therefore, the square root of 676 is 26.

\[OP = 26 \, \text{cm}\]

Conclusion

The value of OP, the distance from the center O to the point P, is 26 cm.

Revision Table: Circle Geometry Formulas

Additional Information on Tangents and Circles

A tangent is a line that touches a circle at exactly one point. This point is called the point of tangency or point of contact (Q in this case). The line segment from an external point (P) to the point of tangency (Q) is called the tangent segment (PQ).

Important points regarding tangents:

• There is only one tangent line to a circle at any given point on the circle.
• From an external point, two tangent segments can be drawn to a circle, and these segments are equal in length.
• The line segment joining the center of the circle to the external point (OP) bisects the angle between the two tangent segments (if two are drawn from P).
• The line segment OP also bisects the angle between the radii drawn to the points of tangency.

Understanding the relationship between the radius and the tangent, as used in this problem, is crucial for solving many geometry questions involving circles.