The equation of the circle of radius 5 and touching the co-ordinate axes in third quadrant is

The equation of the circle of radius 5 and touching the co-ordinate axes in third quadrant is
  1. $(x-5)^2+(y+5)^2=25$
  2. $(x+5)^2+(y+5)^2=25$
  3. $(x+4)^2+(y+4)^2=25$
  4. $(x+6)^2+(y+6)^2=25$

Solution

Since the circle touches the coordinate axes in third quadrant therefore.
centre of the circle is $(-5,-5)$ Then equation of circle is, $(x+5)^2+(y+5)^2=25$.

Asked in: AP EAMCET 2002

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