Question

The equations of two circles which touch the coordinate axes and whose centres lie on the line \[ x-2y=3 \] are

• A. \(x^2+y^2+6x+6y+9=0,\quad x^2+y^2-2x+2y+1=0\)
• B. \(x^2+y^2-6x-6y-9=0,\quad x^2+y^2+2x-2y-1=0\)
• C. \(x^2+y^2-6x+6y+9=0,\quad x^2+y^2+2x+2y-1=0\)
• D. \(x^2+y^2-6x+6y-9=0,\quad x^2+y^2-2x+2y-1=0\)

Hint:

If a circle touches both coordinate axes, then \[ |h|=|k|=r \] where \((h,k)\) is the centre and \(r\) is the radius.

Answer 1

The Correct Option is D

Step 1: What it means to touch both axes.
If a circle touches both coordinate axes, its centre is equally far from the $x$-axis and the $y$-axis, and that common distance is the radius. So the centre looks like $(h,k)$ with $|h|=|k|=r$.

Step 2: Bring in the line condition.
The centre must also lie on $x-2y=3$. Since $|h|=|k|$, the centre is of the form $(a,a)$ or $(a,-a)$. We test both.

Step 3: Try centre $(a,a)$.
Putting $x=a,\ y=a$ in $x-2y=3$ gives $a-2a=3$, so $a=-3$. The centre is $(-3,-3)$ and radius $r=|-3|=3$.

Step 4: Write the first circle.
With centre $(-3,-3)$ and $r=3$: $(x+3)^2+(y+3)^2=9$. Expanding, $x^2+y^2+6x+6y+9=0$.

Step 5: Try centre $(a,-a)$.
Putting $x=a,\ y=-a$ in $x-2y=3$ gives $a+2a=3$, so $a=1$. The centre is $(1,-1)$ and radius $r=1$.

Step 6: Write the second circle and match.
With centre $(1,-1)$ and $r=1$: $(x-1)^2+(y+1)^2=1$, which expands to $x^2+y^2-2x+2y+1=0$. The two circles $x^2+y^2+6x+6y+9=0$ and $x^2+y^2-2x+2y+1=0$ match option four. \[ \boxed{x^2+y^2+6x+6y+9=0,\ \ x^2+y^2-2x+2y+1=0} \]