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BITSAT2021MathematicsCircleActual

The equation of the circle, which touches the line y=5 and passes through (-1,2) and (1,2) is

Options

  1. A9 x²+9 y²-60 y+75=0
  2. B9 x²+9 y²-60 x-75=0
  3. C9 x²+9 y²+60 y-75=0
  4. D9 x²+9 y²+60 x+75=0

Correct answer

A. 9 x²+9 y²-60 y+75=0

Step-by-step solution

The centre of the circle is on the perpendicular bisector of the line joining (-1,2) and (1,2) , which is the y -axis. The ordinate of the centre is given by array l (5-y)²=1+(y-2)² y= 10 3 array Hence, eq. of the circle is: array l x²+ (y- 10 3 )²= ( 5 3 )² 9 x²+9 y²-60 y+75=0 array

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