Question

find the equation of the circle which touches line x+y = 0 at the origin and passes through the point (3,-1)

Answer 1

is it the line that passes through (3,-1) or the circle?

Answer 2

\[(x-p)^2+(y-q)^2=r^2\]\[(0,0)\Rightarrow p^2+q^2=r^2\]\[(3,-1)\Rightarrow (3-p)^2+(-1-q)^2=r^2\]\[((x-p)^2+(y-q)^2)'=(r^2)'\]\[2(x-p)+2(y-q)y'=0\quad\Rightarrow\quad y'=-\frac{x-p}{y-q}\]\[y'(0,0)=-\frac{p}{q}=-1\quad\Rightarrow\quad p=q\]\[2p^2=r^2\quad,\quad (3-p)^2+(-1-p)^2=r^2\]\[(3-p)^2+(-1-p)^2=2p^2\]\[9-6p+p^2+1+2p+p^2=2p^2\]\[10-4p=0\quad\Rightarrow\quad p=q=2.5\quad,\quad r^2=12.5\]\[(x-2.5)^2+(y-2.5)^2=12.5\]