question_answer

The equation of the circle of  radius 5 and  touching the coordinate axes  in third quadrant is   [EAMCET 2002]

A)            \[{{(x-5)}^{2}}+{{(y+5)}^{2}}=25\]                              

B)            \[{{(x+4)}^{2}}+{{(y+4)}^{2}}=25\]

C)            \[{{(x+6)}^{2}}+{{(y+6)}^{2}}=25\]                            

D)            \[{{(x+5)}^{2}}+{{(y+5)}^{2}}=25\]

Correct Answer: D

Solution :

           Since circle touches the co-ordinate axes in III quadrant.                    \[\therefore \] Radius \[=-h=-k\]. Hence \[h=k=-5\]                    \[\therefore \] Equation of circle is \[{{(x+5)}^{2}}+{{(y+5)}^{2}}=25.\]