A variable circle is described to pass through the point (a, 0) and touch the line `x+y=0`. Locus of the centre of the above circle is

Options

A. parabola
B. ellipse
C. hyperbola
D. none of these

Answer

To find the locus of the center of a variable circle that passes through the point (a, 0) and touches the line \(x + y = 0\), we can follow these steps:

### Step 1: Understand the Circle's Properties
The circle passes through the point (a, 0) and touches the line \(x + y = 0\). The center of the circle can be denoted as \(C(h, k)\).

### Step 2: Write the Equation of the Circle
The general equation of a circle with center \(C(h, k)\) and radius \(r\) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

### Step 3: Determine the Radius
Since the circle touches the line \(x + y = 0\), the distance from the center \(C(h, k)\) to the line must equal the radius \(r\). The distance \(d\) from a point \((h, k)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
For the line \(x + y = 0\), we have \(A = 1\), \(B = 1\), and \(C = 0\). Therefore, the distance from the center to the line is:
\[ d = \frac{|h + k|}{\sqrt{1^2 + 1^2}} = \frac{|h + k|}{\sqrt{2}} \]

### Step 4: Use the Circle's Radius
The radius \(r\) of the circle can also be expressed as the distance from the center \(C(h, k)\) to the point (a, 0):
\[ r = \sqrt{(h - a)^2 + (k - 0)^2} = \sqrt{(h - a)^2 + k^2} \]

### Step 5: Set the Distances Equal
Since the circle touches the line, we set the distance from the center to the line equal to the radius:
\[ \frac{|h + k|}{\sqrt{2}} = \sqrt{(h - a)^2 + k^2} \]

### Step 6: Square Both Sides
To eliminate the square root, we square both sides:
\[ \frac{(h + k)^2}{2} = (h - a)^2 + k^2 \]

### Step 7: Expand and Simplify
Expanding both sides gives:
\[ \frac{h^2 + 2hk + k^2}{2} = h^2 - 2ah + a^2 + k^2 \]
Multiplying through by 2 to eliminate the fraction:
\[ h^2 + 2hk + k^2 = 2h^2 - 4ah + 2a^2 + 2k^2 \]
Rearranging terms:
\[ 0 = h^2 - 4ah + a^2 + k^2 - 2hk \]

### Step 8: Rearranging to Find the Locus
This can be rearranged to:
\[ h^2 - 4ah + (k^2 - 2hk + a^2) = 0 \]
This is a quadratic in \(h\). For the locus of the center, we can complete the square or analyze the equation further.

### Final Step: Identify the Locus
The equation represents a parabola in the \(hk\)-plane, which gives us the locus of the center of the circle.