Real numbers satisfying system of equations

CAT 2019 Slot 2 · QA · Hard · Number theory

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Let a,b,x,ya, b, x, y be real numbers such that a2+b2=25a^2+b^2 = 25, x2+y2=169x^2+y^2 = 169 and ax+by=65ax + by = 65. If k=ay−bxk = ay - bx, then

  1. A.

    k=0k = 0

  2. B.

    k>513k > \frac{5}{13}

  3. C.

    k=513k = \frac{5}{13}

  4. D.

    0<k≤5130 < k \le \frac{5}{13}

Answer

A

Explanation

Using the algebraic identity (Lagrange's identity): (a2+b2)(x2+y2)=(ax+by)2+(ay−bx)2(a^2 + b^2)(x^2 + y^2) = (ax + by)^2 + (ay - bx)^2

Substitute the given values: 25×169=652+k225 \times 169 = 65^2 + k^2 4225=4225+k24225 = 4225 + k^2 k2=0  ⟹  k=0k^2 = 0 \implies k = 0

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