Question

Chords AB CD of a circle intersect at right angle at the point P. If the lengths of AP, PB, CP, PD are 2, 6, 3, 4 units respectively, then the radius of the circle is:

• A. \(4\) units
• B. \(\frac{\sqrt{65}}{2}\) units
• C. \(\frac{\sqrt{67}}{2}\) units
• D. \(\frac{\sqrt{66}}{2}\) units

Hint:

When dealing with intersecting chords, use the Intersecting Chords Theorem and the formula for the radius to find the required quantities.

Answer 1

The Correct Option is B

We are given:

\(AP = 2\), \(PB = 6\), \(CP = 3\), \(PD = 4\).

The two chords \(AB\) and \(CD\) intersect at right angles at point \(P\).

Step 1: Use Geometry of Intersecting Chords.

The formula for the radius \(r\) of the circle when two chords intersect perpendicularly is:

\[ r^2 = \frac{AP^2 + PB^2 + CP^2 + PD^2}{2}. \]

Step 2: Substitute the Given Values.

Substitute the lengths of \(AP\), \(PB\), \(CP\), and \(PD\):

\[ r^2 = \frac{2^2 + 6^2 + 3^2 + 4^2}{2}. \]

Simplify:

\[ r^2 = \frac{4 + 36 + 9 + 16}{2}. \] \[ r^2 = \frac{65}{2}. \]

Step 3: Calculate the Radius.

Take the square root of both sides:

\[ r = \sqrt{\frac{65}{2}} = \frac{\sqrt{65}}{2}. \]

Conclusion: The radius of the circle is:

\[ \frac{\sqrt{65}}{2} \, \text{units}. \]