Question

A circle passes through the point \((0,1)\) and touches the parabola \(y = x^2\) at the point \((1,1)\). The centre of the circle is...

• A. \((-\frac{1}{2},-\frac{5}{2})\)
• B. \((\frac{1}{2},-\frac{5}{2})\)
• C. \((\frac{1}{2},\frac{5}{4})\)
• D. \((-\frac{1}{2},\frac{5}{4})\)

Hint:

The centre lies on the normal to the parabola at (1,1) and on the perpendicular bisector of the chord to (0,1).

Answer 1

The Correct Option is C

Step 1: Understand the setup
The circle passes through \((0,1)\) and \((1,1)\), and it touches \(y = x^2\) at \((1,1)\). So the radius to \((1,1)\) is perpendicular to the tangent of the parabola there.

Step 2: Normal to the parabola
\(\frac{dy}{dx} = 2x = 2\) at \(x = 1\), so the tangent slope is 2 and the normal slope is \(-\frac{1}{2}\). The centre \((h, k)\) lies on \(y - 1 = -\frac{1}{2}(x - 1)\), so \(k - 1 = -\frac{1}{2}(h - 1)\).

Step 3: Use the two points
The centre is equidistant from \((0,1)\) and \((1,1)\): \(h^2 + (k-1)^2 = (h-1)^2 + (k-1)^2\). So \(h^2 = h^2 - 2h + 1\), giving \(h = \frac{1}{2}\).

Step 4: Find k
\(k - 1 = -\frac{1}{2}\left(\frac{1}{2} - 1\right) = \frac{1}{4}\), so \(k = \frac{5}{4}\). The centre is \(\left(\frac{1}{2}, \frac{5}{4}\right)\), option (C). The options with \(-\frac{5}{2}\) place the centre far below both points and would not give equal distances to both.

Final Answer:
The centre is (1/2, 5/4). This is option (C). \[ \boxed{\text{(C) }\left(\frac{1}{2},\frac{5}{4}\right)} \]