AP EAMCET2007MathematicsCircle
The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines x=0 and y=0 is
Options
- Ax^2+y^2-6 x+6 y+9=0
- Bx^2+y^2-6 x-6 y+9=0
- Cx^2+y^2+6 x-6 y+9=0
- Dx^2+y^2+6 x+6 y+9=0
Correct answer
A. x^2+y^2-6 x+6 y+9=0
Step-by-step solution
Given, radius =3 , Since, the circle touching both the coordinate axes in 4th quadrant, so equation is gathered (x-3)^2+(y+3)^2=3^2 x^2+9-6 x+y^2+9+6 y=9 x^2+y^2-6 x+6 y+9=0 gathered