Question:
The circle possessing \(y\)-axis as its tangent at \((0,2)\) and passing through \((-1,0)\), also passes through
The circle possessing \(y\)-axis as its tangent at \((0,2)\) and passing through \((-1,0)\), also passes through
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If a circle touches an axis, its centre lies on the perpendicular line through the point of contact.
Updated On: Jun 25, 2026
- \(\left(-\frac{3}{2},0\right)\)
- \(\left(-\frac{5}{2},2\right)\)
- \(\left(-\frac{3}{2},\frac{5}{2}\right)\)
- \((-4,0)\)
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The Correct Option is D
Solution and Explanation
Step 1: Assume the centre of the circle.
Since the \(y\)-axis is tangent to the circle at \((0,2)\), the radius at the point of contact is perpendicular to the \(y\)-axis.
So, the centre lies on the horizontal line passing through \((0,2)\).
Let the centre be \[ (a,2) \] Then radius is \[ |a| \]
Step 2: Use the point \((-1,0)\).
Since the circle passes through \((-1,0)\), \[ (-1-a)^2+(0-2)^2=a^2 \] \[ (-1-a)^2+4=a^2 \] \[ (a+1)^2+4=a^2 \] \[ a^2+2a+1+4=a^2 \] \[ 2a+5=0 \] \[ a=-\frac{5}{2} \] So, the centre is \[ \left(-\frac{5}{2},2\right) \]
Step 3: Find the other point on the \(x\)-axis.
The radius is \[ \frac{5}{2} \] Equation of the circle is \[ \left(x+\frac{5}{2}\right)^2+(y-2)^2=\left(\frac{5}{2}\right)^2 \] For points on the \(x\)-axis, \[ y=0 \] So, \[ \left(x+\frac{5}{2}\right)^2+4=\frac{25}{4} \] \[ \left(x+\frac{5}{2}\right)^2=\frac{9}{4} \] \[ x+\frac{5}{2}=\pm \frac{3}{2} \] Thus, \[ x=-1 \] or \[ x=-4 \] Since \((-1,0)\) is already given, the other point is \[ (-4,0) \]
Step 4: Final conclusion.
Therefore, the circle also passes through \[ \boxed{(-4,0)} \]
Since the \(y\)-axis is tangent to the circle at \((0,2)\), the radius at the point of contact is perpendicular to the \(y\)-axis.
So, the centre lies on the horizontal line passing through \((0,2)\).
Let the centre be \[ (a,2) \] Then radius is \[ |a| \]
Step 2: Use the point \((-1,0)\).
Since the circle passes through \((-1,0)\), \[ (-1-a)^2+(0-2)^2=a^2 \] \[ (-1-a)^2+4=a^2 \] \[ (a+1)^2+4=a^2 \] \[ a^2+2a+1+4=a^2 \] \[ 2a+5=0 \] \[ a=-\frac{5}{2} \] So, the centre is \[ \left(-\frac{5}{2},2\right) \]
Step 3: Find the other point on the \(x\)-axis.
The radius is \[ \frac{5}{2} \] Equation of the circle is \[ \left(x+\frac{5}{2}\right)^2+(y-2)^2=\left(\frac{5}{2}\right)^2 \] For points on the \(x\)-axis, \[ y=0 \] So, \[ \left(x+\frac{5}{2}\right)^2+4=\frac{25}{4} \] \[ \left(x+\frac{5}{2}\right)^2=\frac{9}{4} \] \[ x+\frac{5}{2}=\pm \frac{3}{2} \] Thus, \[ x=-1 \] or \[ x=-4 \] Since \((-1,0)\) is already given, the other point is \[ (-4,0) \]
Step 4: Final conclusion.
Therefore, the circle also passes through \[ \boxed{(-4,0)} \]
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