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(A) Since the circle touches both coordinate axes,its centre $(h, k)$ must satisfy $|h| = |k| = r$,where $r$ is the radius.Thus,the centre is either $

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$x^2+y^2-2x+2y+1=0$

Vedclass · Answer

(A) Since the circle touches both coordinate axes,its centre $(h, k)$ must satisfy $|h| = |k| = r$,where $r$ is the radius.Thus,the centre is either $(r, r), (r, -r), (-r, r),$ or $(-r, -r)$.The centre lies on the line $x-2y-3=0$.Case $1$: Centre is $(r, r)$. Then $r-2r-3=0$ $\Rightarrow -r=3$ $\Rightarrow r=-3$. Since $r > 0$,this is not possible.Case $2$: Centre is $(r, -r)$. Then $r-2(-r)-3=0$ $\Rightarrow 3r=3$ $\Rightarrow r=1$. The centre is $(1, -1)$.The equation of the circle is $(x-1)^2+(y+1)^2=1^2$ $\Rightarrow x^2-2x+1+y^2+2y+1=1$ $\Rightarrow x^2+y^2-2x+2y+1=0$.Case $3$: Centre is $(-r, r)$. Then $-r-2r-3=0$ $\Rightarrow -3r=3$ $\Rightarrow r=-1$. Not possible.Case $4$: Centre is $(-r, -r)$. Then $-r-2(-r)-3=0 \Rightarrow r=3$. The centre is $(-3, -3)$.The equation of the circle is $(x+3)^2+(y+3)^2=3^2$ $\Rightarrow x^2+6x+9+y^2+6y+9=9$ $\Rightarrow x^2+y^2+6x+6y+9=0$ (not in options).Therefore,the correct equation is $x^2+y^2-2x+2y+1=0$.

The circle touching the coordinate axes with its centre lying on $x-2y-3=0$ is

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