AP EAMCET · Maths · Straight Lines
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
- A 2
- B 3
- C 4
- D 5
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
The intersection point of the curve \(x^2+y^2=4\) with \(x+y=a\), where \((a>0)\) \(x^2+(a-x)^2=4\) \(x^2+a^2+x^2-2 a x=4\) \(2 x^2-2 a x+\left(a^2-4\right)=0\) \(x^2-a x+\left(\frac{a^2}{2}-2\right)=0\) \(x=\frac{+a \pm \sqrt{a^2-4\left(\frac{a^2}{2}-2\right)}}{2}\)…
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