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

Given:

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

Chords \(AB\) and \(CD\) intersect at right angles at \(P\).

Step 1: Apply Intersecting Chords Geometry.

The radius \(r\) can be found using the formula:

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

Step 2: Input the Values.

Substitute \(AP\), \(PB\), \(CP\), 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: Compute the Radius.

Take the square root:

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

Answer: The circle's radius is:

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