InterviewSolution
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InterviewSolution
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Equation of a circle which passes through (3, 6) and touches the axes isA. x2 + y2 + 6x + 6y + 3 = 0B. x2 + y2 – 6x – 6y – 9 = 0C. x2 + y2 – 6x – 6y + 9 = 0D. none of these |
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Answer» When circle touches both the axes, the co – ordinates of the centre and its radius are equal in their magnitude, h = k – r Since, the equation of a circle having centre (h,k), having radius as "r" units, is (x – h)2 + (y – k)2 = r2 (3 – h)2 + (6 – h)2 = h2 9 + h2 - 6h + 36 + h2 - 12h = h2 h2 - 18h + 45 = 0 h2 - 15h – 3h + 45 = 0 h (h – 15) – 3 (h – 15) = 0 (h – 3) (h – 15) = 0 h = 3 or h = 15 Co – ordinates of centre are (3, 3) or (15, 15) (x – h)2 + (y – k)2 = r2 Equation, having centre (3, 3) (x – 3)2 + (y – 3)2 = 32 x2 - 6x + 9 + y2 - 6y + 9 – 9 = 0 x2 - 6x + y2 - 6y + 9 = 0 Equation, having centre (15, 15) (x – 15)2 + (y – 15)2 = 152 x2 - 30x + 225 + y2 - 30y + 225 – 225 = 0 x2 - 30x + y2 - 30y + 225 = 0 Hence the equations are x2 - 6x + y2 - 6y + 9 = 0 or x2 - 30x + y2 - 30y + 225 = 0. Option (C) is the answer. |
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