The equation of the circle whose radius is 3 and which touches the circle x 2 + y 2 – 4x…

Mathematics · JEE Advanced · NTA ExamsCo-ordinate Geometry

The equation of the circle whose radius is 3 and which touches the circle x2 + y2 – 4x – 6y – 12 = 0 internally at the point (–1, –1) is-
  1. \(\left(x-\frac{4}{5}\right)^{2}+\left(y+\frac{7}{5}\right)^{2}=3^{2}\)
  2. \(\left(x-\frac{4}{5}\right)^{2}+\left(y-\frac{7}{5}\right)^{2}=3^{2}\)
  3. (x–8)2 + (y–1)2 = 32
  4. None of these

Answer

(B) (x- 4 5 )^ 2 + (y- 7 5 )^ 2 =3^ 2

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