Question

Find the equation of the circle in the third quadrant having radius $3$ and touching both the axes.

Answer 1

Let radius be $r$, here $r=3$

And center be $(h,k)$

Since circle touches axes in third quadrant,

So, $(h,k)=(-r,-r)$

So, center$=(-3,-3)$

Thus equation will be

$(x+3{)}^2+(y+3{)}^2=3^2$

So, $(x+3{)}^2+(y+3{)}^2=9$