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BITSAT 2016
Exam
BITSAT
Year
2016
Subject
Mathematics
Chapter
Circle

BITSAT 2016 · Mathematics · Circle

The length of the chord x+y=3 intercepted by the circle x^ 2 +y^ 2...

The length of the chord x+y=3x+y=3 intercepted by the circle x2+y2−2x−2y−2=0x^{2}+y^{2}-2 x-2 y-2=0 is

Options

  1. A72\frac{7}{2}
  2. B332\frac{3 \sqrt{3}}{2}
  3. C14\sqrt{14}
  4. D72\frac{\sqrt{7}}{2}

Correct answer

C. 14\sqrt{14}

Step-by-step solution

The centre of the circle is C(1,1))\mathrm{C}(1,1)) and radiusofthecircleis2,perpendiculardistancefromCradius of the circle is 2 , perpendicular distance from \mathrm{C} on AB\mathrm{AB}, the chord x+y=3)(\mathrm{x}+\mathrm{y}=3) (CD=∣1+1−32∣=12 and ⇒AD=4−12=72[AD=AC2−CD2]\begin{aligned} \mathrm{CD}=\left|\frac{1+1-3}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}\\ \text{ and } \Rightarrow \mathrm{AD}=\sqrt{4-\frac{1}{2}}=\sqrt{\frac{7}{2}}\left[\mathrm{AD}=\sqrt{\mathrm{AC}^{2}-\mathrm{CD}^{2}}\right] \end{aligned})Hence,thelengthofthechordAB=2AD=272=14) Hence, the length of the chord \mathrm{AB}=2 \mathrm{AD}=2 \sqrt{\frac{7}{2}}=\sqrt{14}.
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