Question

PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

• A. \(a^2 + (a + 2)^2 = (2b)^2\)
• B. \(b^2 = a + 4\)
• C. \(2a^2 + 1 = b^2\)
• D. \(b^2 = a + 1\)

Hint:

Always draw the radius to the point of contact of a tangent to identify the right-angled triangle and apply the Pythagorean property.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
A tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, the triangle formed by the center, the point of contact, and an external point is a right-angled triangle.
Step 2: Key Formula or Approach:
In \(\triangle OQP\), since \(PQ\) is tangent at \(Q\), \(\angle OQP = 90^{\circ}\).
By Pythagoras Theorem:
\[ OQ^2 + PQ^2 = OP^2 \]
Step 3: Detailed Explanation:
Given:
\(OQ = a\)
\(PQ = 2b\)
\(OP = a + 2\)
Applying Pythagoras Theorem:
\[ a^2 + (2b)^2 = (a + 2)^2 \]
\[ a^2 + 4b^2 = a^2 + 4a + 4 \]
Subtracting \(a^2\) from both sides:
\[ 4b^2 = 4a + 4 \]
Dividing the entire equation by \(4\):
\[ b^2 = a + 1 \]
Step 4: Final Answer:
The relation between \(a\) and \(b\) is \(b^2 = a + 1\).