Question:
The centre of the circle that passes through the point \((0,1)\) and touches the curve \(y=x^2\) at \((2,4)\) is
The centre of the circle that passes through the point \((0,1)\) and touches the curve \(y=x^2\) at \((2,4)\) is
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If a circle touches a curve at a point, then the centre of the circle lies on the normal to the curve at that point.
Updated On: Jun 25, 2026
- \(\left(-\frac{16}{5},\frac{27}{10}\right)\)
- \(\left(-\frac{16}{7},\frac{53}{10}\right)\)
- \(\left(-\frac{16}{5},\frac{53}{10}\right)\)
- \(\left(-\frac{16}{5},-\frac{53}{10}\right)\)
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The Correct Option is C
Solution and Explanation
Step 1: Let the centre of the circle be \((h,k)\).
The circle touches the curve \[ y=x^2 \] at the point \[ (2,4) \] So, the centre of the circle must lie on the normal to the curve at \((2,4)\).
Step 2: Find the slope of tangent and normal.
For \[ y=x^2, \] we have \[ \frac{dy}{dx}=2x \] At \(x=2\), \[ \frac{dy}{dx}=4 \] So, slope of tangent is \[ 4 \] Therefore, slope of normal is \[ -\frac{1}{4} \]
Step 3: Equation of normal at \((2,4)\).
Using point-slope form: \[ y-4=-\frac{1}{4}(x-2) \] Since the centre \((h,k)\) lies on this normal, \[ k-4=-\frac{1}{4}(h-2) \] Multiplying by \(4\), \[ 4k-16=-h+2 \] \[ h+4k-18=0 \] So, \[ h+4k=18 \]
Step 4: Use equal radius condition.
The circle passes through \((0,1)\) and touches the curve at \((2,4)\).
Therefore, distance from centre \((h,k)\) to \((0,1)\) is equal to distance from centre \((h,k)\) to \((2,4)\).
So, \[ (h-0)^2+(k-1)^2=(h-2)^2+(k-4)^2 \] Expanding, \[ h^2+k^2-2k+1=h^2-4h+4+k^2-8k+16 \] Cancelling common terms, \[ -2k+1=-4h-8k+20 \] \[ 4h+6k-19=0 \] So, \[ 4h+6k=19 \]
Step 5: Solve the two equations.
We have \[ h+4k=18 \] and \[ 4h+6k=19 \] From the first equation, \[ h=18-4k \] Substitute in the second equation: \[ 4(18-4k)+6k=19 \] \[ 72-16k+6k=19 \] \[ 72-10k=19 \] \[ -10k=-53 \] \[ k=\frac{53}{10} \] Now, \[ h=18-4\left(\frac{53}{10}\right) \] \[ h=18-\frac{106}{5} \] \[ h=\frac{90-106}{5} \] \[ h=-\frac{16}{5} \]
Step 6: Final conclusion.
Therefore, the centre of the circle is \[ \boxed{\left(-\frac{16}{5},\frac{53}{10}\right)} \]
The circle touches the curve \[ y=x^2 \] at the point \[ (2,4) \] So, the centre of the circle must lie on the normal to the curve at \((2,4)\).
Step 2: Find the slope of tangent and normal.
For \[ y=x^2, \] we have \[ \frac{dy}{dx}=2x \] At \(x=2\), \[ \frac{dy}{dx}=4 \] So, slope of tangent is \[ 4 \] Therefore, slope of normal is \[ -\frac{1}{4} \]
Step 3: Equation of normal at \((2,4)\).
Using point-slope form: \[ y-4=-\frac{1}{4}(x-2) \] Since the centre \((h,k)\) lies on this normal, \[ k-4=-\frac{1}{4}(h-2) \] Multiplying by \(4\), \[ 4k-16=-h+2 \] \[ h+4k-18=0 \] So, \[ h+4k=18 \]
Step 4: Use equal radius condition.
The circle passes through \((0,1)\) and touches the curve at \((2,4)\).
Therefore, distance from centre \((h,k)\) to \((0,1)\) is equal to distance from centre \((h,k)\) to \((2,4)\).
So, \[ (h-0)^2+(k-1)^2=(h-2)^2+(k-4)^2 \] Expanding, \[ h^2+k^2-2k+1=h^2-4h+4+k^2-8k+16 \] Cancelling common terms, \[ -2k+1=-4h-8k+20 \] \[ 4h+6k-19=0 \] So, \[ 4h+6k=19 \]
Step 5: Solve the two equations.
We have \[ h+4k=18 \] and \[ 4h+6k=19 \] From the first equation, \[ h=18-4k \] Substitute in the second equation: \[ 4(18-4k)+6k=19 \] \[ 72-16k+6k=19 \] \[ 72-10k=19 \] \[ -10k=-53 \] \[ k=\frac{53}{10} \] Now, \[ h=18-4\left(\frac{53}{10}\right) \] \[ h=18-\frac{106}{5} \] \[ h=\frac{90-106}{5} \] \[ h=-\frac{16}{5} \]
Step 6: Final conclusion.
Therefore, the centre of the circle is \[ \boxed{\left(-\frac{16}{5},\frac{53}{10}\right)} \]
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