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The equation of the circle which passes through the point \((4,5)\) and its centre at \((2,2)\) is

Question and options

The equation of the circle which passes through the point (4,5)(4,5) and its centre at (2,2)(2,2) is
  1. (x−2)+(y−2)=13(x-2)+(y-2)=13
  2. (x−2)2+(y−2)2=13(x-2)^{2}+(y-2)^{2}=13
  3. x2+y2=13x^{2}+y^{2}=13
  4. (x−4)2+(y−5)2=13(x-4)^{2}+(y-5)^{2}=13

Answer and explanation

Correct answer: B - \((x-2)^{2}+(y-2)^{2}=13\)
r^2 = (4-2)^2 + (5-2)^2 = 13, so the circle is (x-2)^2 + (y-2)^2 = 13.

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