Question

The locus of the centre of the circle touching the x-axis and passing through the point \((-1,1)\) is

• A. a circle with centre at \((-1, \frac{1}{2})\)
• B. a pair of lines intersecting at \((-1, 1)\)
• C. a parabola with focus at \((-1,1)\)
• D. a hyperbola with centre at \((-1,1)\)

Hint:

By definition, a parabola is the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix). Here, the center is equidistant from \(P(-1,1)\) and the line \(y=0\) (x-axis) if we consider \(y=0\) as the tangent but actually the condition \(CP = \text{dist}(C, \text{x-axis})\) matches the parabola definition perfectly with \(P\) as focus and x-axis as directrix.

Answer 1

The Correct Option is C

Step 1: Define the Circle's Properties:
Let the centre of the circle be \(C(h, k)\). Since the circle touches the x-axis, its radius \(r\) is equal to the absolute value of the y-coordinate of the centre. Thus, \(r = |k|\).
Step 2: Use the Distance Condition:
The circle passes through the point \(P(-1, 1)\). Therefore, the distance from the centre \(C(h, k)\) to \(P(-1, 1)\) must equal the radius \(r\). \[ \sqrt{(h - (-1))^2 + (k - 1)^2} = |k| \] Square both sides: \[ (h + 1)^2 + (k - 1)^2 = k^2 \]
Step 3: Simplify the Equation:
\[ (h + 1)^2 + k^2 - 2k + 1 = k^2 \] \[ (h + 1)^2 - 2k + 1 = 0 \] \[ (h + 1)^2 = 2k - 1 \] \[ (h + 1)^2 = 2\left(k - \frac{1}{2}\right) \]
Step 4: Identify the Locus:
Replacing \((h, k)\) with \((x, y)\), the locus is: \[ (x + 1)^2 = 2\left(y - \frac{1}{2}\right) \] This equation represents a parabola of the form \((X)^2 = 4aY\), where \(X = x+1\) and \(Y = y - \frac{1}{2}\). Here, \(4a = 2 \implies a = \frac{1}{2}\).
Step 5: Find the Focus:
The vertex of the parabola is at \((-1, \frac{1}{2})\). The focus is located at a distance \(a\) above the vertex (since the parabola opens upwards). Focus coordinate \( = (-1, \frac{1}{2} + a) = (-1, \frac{1}{2} + \frac{1}{2}) = (-1, 1) \). Thus, the locus is a parabola with its focus at \((-1, 1)\).