A(a→), B(b→), C(c→), D(d→) are four concyclic points, such that xa→+yb→+zc→+td→=0→, x+y+z+t=0, where x, y, z, t are constants not all zero. If the chords AB and CD intersect at P, then

TS EAMCET · Maths · Vector Algebra

$A (\vec{a}), B (\vec{b}), C (\vec{c}), D (\vec{d})$ are four concyclic points, such that $x \vec{a} + y \vec{b} + z \vec{c} + t \vec{d} = \vec{0}, x + y + z + t = 0$ , where $x, y, z, t$ are constants not all zero. If the chords $A B$ and $C D$ intersect at $P$ , then

A $|x y| {|\vec{a} + \vec{c}|}^{2} = |z t| {|\vec{b} + \vec{d}|}^{2}$
B $|x y| {|\vec{a} - \vec{b}|}^{2} = |z t| {|\vec{c} - \vec{d}|}^{2}$
C $|x t| {|\vec{a} - \vec{c}|}^{2} = |y z| {|\vec{b} - \vec{d}|}^{2}$
D $|x z| {|\vec{b} + \vec{d}|}^{2} = |y t| {|\vec{a} + \vec{c}|}^{2}$

Verified Solution

Answer & Solution

Correct Answer

(B) $|x y| {|\vec{a} - \vec{b}|}^{2} = |z t| {|\vec{c} - \vec{d}|}^{2}$

Step-by-step Solution

Detailed explanation

We are given that Aa→, Bb→, Cc→ & Dd→ are four concyclic points such that xa→+yb→+zc→+td→=0→ ...i x+y+z+t=0 ...ii Now, we know that angle subtended by chord on same side of circle…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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