Q57 / 160 · AP EAMCET 2010 · Mathematics

A pair of perpendicular lines passes through the origin and also through the points of intersection of the curve $x^2+y^2=4$ with $x+y=a$, where $a>0$. Then $a$ is equal to

  • Correct answer
    A
    2
  • B
    3
  • C
    4
  • D
    5

Solution

The intersection point of the curve $x^2+y^2=4$ with $x+y=a$, where $(a>0)$ $x^2+(a-x)^2=4$ $x^2+a^2+x^2-2 a x=4$ $2 x^2-2 a x+\left(a^2-4\right)=0$ $x^2-a x+\left(\frac{a^2}{2}-2\right)=0$ $x=\frac{+a \pm \sqrt{a^2-4\left(\frac{a^2}{2}-2\right)}}{2}$ $x=\frac{a \pm \sqrt{a^2-2 a^2+8}}{2}=\frac{a \pm \sqrt{8-a^2}}{2}$ Since, here the point of intersection should be real number, iff $8-a^2 \geq 0$ ie, $\quad a^2 \leq 8, a \leq 2 \sqrt{2}$ $a \leq 2.82$ Hence, the value is $a=2$ according to option.

Hence, the correct answer is option: 'A'.

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