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Vector Algebra

Question
CBSEENMA12036020
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If one of the diameters of the circle, given by the equation, x2+y2−4x+6y−12=0, is a chord of a circle S, whose centre is at (−3, 2), then the radius of S is:

  • 5 square root of 2
  • 5 square root of 3

  • 5

  • 10

Solution

B.

5 square root of 3

Given equation of a circle is x2 + y2 -4x +6y -12 = 0, whose centre is (2,-3) and radius
space equals space square root of 2 squared plus left parenthesis negative 3 right parenthesis squared plus 12 end root equals square root of 4 plus 9 plus 12 space equals space 5 end root
Now, according to given information, we have the following figure.

x2+y2-4x +6y-12 =0
Clearly, AO perpendicular to BC, as O is mid-point of the chord.
Now in ΔAOB,. we have
OA space equals space square root of left parenthesis negative 3 minus 2 right parenthesis squared plus left parenthesis 2 plus 3 right parenthesis squared end root equals space square root of 25 plus 25 end root space equals space square root of 50 space end root space equals space 5 square root of 2 and space OB space equals space 5 therefore comma OB space equals space 5 AB space equals space square root of OA squared plus OB squared end root space equals space square root of 50 plus 25 end root space equals space square root of 75 space equals space 5 square root of 3

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