Question

If \(ABCD\) is a cyclic quadrilateral with \(R\) as the radius of the circumcircle and \[ (AB)^2+(CD)^2=4R^2, \] then

• A. \(\vec{b}\cdot\vec{c}-\vec{a}\cdot\vec{d}=0\)
• B. \(\vec{a}\cdot\vec{c}-\vec{b}\cdot\vec{d}=0\)
• C. \(\vec{a}\cdot\vec{b}+\vec{c}\cdot\vec{d}=0\)
• D. \(\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{d}=0\)

Hint:

For points on a circle centered at the origin, use \[ |\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b} \] to convert chord-length conditions into dot-product relations.

Answer 1

The Correct Option is C

Step 1: Use position vectors on the circumcircle.
Let the circumcircle have radius \(R\) and center at the origin. Then \[ |\vec{a}|=|\vec{b}|=|\vec{c}|=|\vec{d}|=R \] Also, \[ AB^2=|\vec{a}-\vec{b}|^2 \] \[ =|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b} \] \[ =2R^2-2\vec{a}\cdot\vec{b} \] Similarly, \[ CD^2=2R^2-2\vec{c}\cdot\vec{d} \]

Step 2: Apply the given condition.
Given \[ AB^2+CD^2=4R^2 \] Substituting, \[ (2R^2-2\vec{a}\cdot\vec{b})+(2R^2-2\vec{c}\cdot\vec{d}) =4R^2 \] \[ 4R^2-2(\vec{a}\cdot\vec{b}+\vec{c}\cdot\vec{d}) =4R^2 \] \[ \vec{a}\cdot\vec{b}+\vec{c}\cdot\vec{d}=0 \]

Step 3: Final conclusion.
\[ \boxed{\vec{a}\cdot\vec{b}+\vec{c}\cdot\vec{d}=0} \]