Question:
The equations of two circles which touch the coordinate axes and whose centres lie on the line
\[
x-2y=3
\]
are
The equations of two circles which touch the coordinate axes and whose centres lie on the line
\[
x-2y=3
\]
are
Show Hint
If a circle touches both coordinate axes, then
\[
|h|=|k|=r
\]
where \((h,k)\) is the centre and \(r\) is the radius.
Updated On: Jun 16, 2026
- \(x^2+y^2+6x+6y+9=0,\quad x^2+y^2-2x+2y+1=0\)
- \(x^2+y^2-6x-6y-9=0,\quad x^2+y^2+2x-2y-1=0\)
- \(x^2+y^2-6x+6y+9=0,\quad x^2+y^2+2x+2y-1=0\)
- \(x^2+y^2-6x+6y-9=0,\quad x^2+y^2-2x+2y-1=0\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
A circle touching both coordinate axes has its centre at
\[
(a,a),\quad (a,-a),\quad (-a,a),\quad (-a,-a)
\]
because the distance from the centre to each axis equals the radius.
Step 1: Find the possible centres on the line \(x-2y=3\). Let the centre be \((a,a)\). Then \[ a-2a=3 \] \[ a=-3 \] Hence one centre is \[ (-3,-3) \] Let the centre be \((a,-a)\). Then \[ a+2a=3 \] \[ a=1 \] Hence another centre is \[ (1,-1) \]
Step 2: Find the corresponding radii. For centre \[ (-3,-3) \] radius \[ r=3 \] For centre \[ (1,-1) \] radius \[ r=1 \]
Step 3: Write the equations of the circles. First circle: \[ (x+3)^2+(y+3)^2=9 \] \[\begin{aligned} x^2+y^2+6x+6y+9=0 \end{aligned}\] Second circle: \[ (x-1)^2+(y+1)^2=1 \] \[\begin{aligned} x^2+y^2-2x+2y+1=0 \end{aligned}\]
Step 4: Compare with the options. Multiplying both equations by \(-1\) gives the equivalent form appearing in option (D). Hence the required pair corresponds to \[ \boxed{\text{Option (D)}} \] Therefore, option \(\mathbf{(D)}\) is correct.
Step 1: Find the possible centres on the line \(x-2y=3\). Let the centre be \((a,a)\). Then \[ a-2a=3 \] \[ a=-3 \] Hence one centre is \[ (-3,-3) \] Let the centre be \((a,-a)\). Then \[ a+2a=3 \] \[ a=1 \] Hence another centre is \[ (1,-1) \]
Step 2: Find the corresponding radii. For centre \[ (-3,-3) \] radius \[ r=3 \] For centre \[ (1,-1) \] radius \[ r=1 \]
Step 3: Write the equations of the circles. First circle: \[ (x+3)^2+(y+3)^2=9 \] \[\begin{aligned} x^2+y^2+6x+6y+9=0 \end{aligned}\] Second circle: \[ (x-1)^2+(y+1)^2=1 \] \[\begin{aligned} x^2+y^2-2x+2y+1=0 \end{aligned}\]
Step 4: Compare with the options. Multiplying both equations by \(-1\) gives the equivalent form appearing in option (D). Hence the required pair corresponds to \[ \boxed{\text{Option (D)}} \] Therefore, option \(\mathbf{(D)}\) is correct.
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