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NDA 2024 Maths · Mathematics · Circles · medium

The equation of the circle having the points (1,2) and (7,10) as the end points of a diameter is

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The equation of the circle having the points (1,2)(1,2) and (7,10)(7,10) as the end points of a diameter is
  1. A. x2+y2+8x+12y+27=0x^2+y^2+8x+12y+27=0
  2. B. x2+y2−8x−12y−27=0x^2+y^2-8x-12y-27=0
  3. C. x2+y2−8x−12y+27=0x^2+y^2-8x-12y+27=0
  4. D. x2+y2−6x−8y+27=0x^2+y^2-6x-8y+27=0
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Answer: (C) x2+y2−8x−12y+27=0x^2+y^2-8x-12y+27=0

Use the diameter form (x−x1)(x−x2)+(y−y1)(y−y2)=0(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0: (x−1)(x−7)+(y−2)(y−10)=0(x-1)(x-7)+(y-2)(y-10)=0, i.e. x2−8x+7+y2−12y+20=0x^2-8x+7+y^2-12y+20=0, which is x2+y2−8x−12y+27=0x^2+y^2-8x-12y+27=0. Trap: a sign error in the linear terms gives the centre (−4,−6)(-4,-6) instead of (4,6)(4,6), and −27-27 results from subtracting the products 77 and 2020.
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