Question:
The circle touching the \(y\)-axis at a distance \(4\) units from the origin and cutting off an intercept \(6\) from \(x\)-axis is:
The circle touching the \(y\)-axis at a distance \(4\) units from the origin and cutting off an intercept \(6\) from \(x\)-axis is:
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If a circle touches an axis, the perpendicular distance of its centre from that axis is equal to its radius.
Updated On: Jun 24, 2026
- \(x^2+y^2+10x-8y+16=0\)
- \(x^2+y^2+5x-8y+16=0\)
- \(x^2+y^2+5x-2y-8=0\)
- \(x^2+y^2+2x-y-12=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the condition of touching the \(y\)-axis.
Since the circle touches the \(y\)-axis at a distance \(4\) units from the origin, the point of contact is \[ (0,4) \] The centre of the circle must lie on the horizontal line through this point.
So, let the centre be \[ (-r,4) \] Since the circle touches the \(y\)-axis, its radius is \[ r \]
Step 2: Write the equation of the circle.
The equation of the circle is \[ (x+r)^2+(y-4)^2=r^2 \] Expanding, \[ x^2+2rx+r^2+y^2-8y+16=r^2 \] \[ x^2+y^2+2rx-8y+16=0 \]
Step 3: Use the \(x\)-axis intercept condition.
The circle cuts an intercept of length \(6\) on the \(x\)-axis.
Put \[ y=0 \] Then, \[ x^2+2rx+16=0 \] For a quadratic equation, the distance between roots is \[ \frac{\sqrt{D}}{|a|} \] Here, \[ D=(2r)^2-4(1)(16) \] \[ D=4r^2-64 \] The length of intercept is \(6\), so \[ \sqrt{4r^2-64}=6 \] Squaring both sides, \[ 4r^2-64=36 \] \[ 4r^2=100 \] \[ r^2=25 \] \[ r=5 \]
Step 4: Substitute \(r=5\).
The equation becomes \[ x^2+y^2+2(5)x-8y+16=0 \] \[ x^2+y^2+10x-8y+16=0 \]
Step 5: Final conclusion.
Hence, the required circle is \[ \boxed{x^2+y^2+10x-8y+16=0} \]
Since the circle touches the \(y\)-axis at a distance \(4\) units from the origin, the point of contact is \[ (0,4) \] The centre of the circle must lie on the horizontal line through this point.
So, let the centre be \[ (-r,4) \] Since the circle touches the \(y\)-axis, its radius is \[ r \]
Step 2: Write the equation of the circle.
The equation of the circle is \[ (x+r)^2+(y-4)^2=r^2 \] Expanding, \[ x^2+2rx+r^2+y^2-8y+16=r^2 \] \[ x^2+y^2+2rx-8y+16=0 \]
Step 3: Use the \(x\)-axis intercept condition.
The circle cuts an intercept of length \(6\) on the \(x\)-axis.
Put \[ y=0 \] Then, \[ x^2+2rx+16=0 \] For a quadratic equation, the distance between roots is \[ \frac{\sqrt{D}}{|a|} \] Here, \[ D=(2r)^2-4(1)(16) \] \[ D=4r^2-64 \] The length of intercept is \(6\), so \[ \sqrt{4r^2-64}=6 \] Squaring both sides, \[ 4r^2-64=36 \] \[ 4r^2=100 \] \[ r^2=25 \] \[ r=5 \]
Step 4: Substitute \(r=5\).
The equation becomes \[ x^2+y^2+2(5)x-8y+16=0 \] \[ x^2+y^2+10x-8y+16=0 \]
Step 5: Final conclusion.
Hence, the required circle is \[ \boxed{x^2+y^2+10x-8y+16=0} \]
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