Question

A circle touches both the y-axis and the line \( x + y = 0 \). Then the locus of its center is:

• A. \( y = \sqrt{2}x \)
• B. \( x = \sqrt{2}y \)
• C. \( y^2 - x^2 = 2xy \)
• D. \( x^2 - y^2 = 2xy \)

Hint:

In problems involving a circle touching a line or an axis, use the distance formula to equate the distance from the center to the line and from the center to the axis, and then simplify the resulting equation to find the locus.

Answer 1

The Correct Option is D

Step 1: Let the center of the circle be at \( (h, k) \) and its radius be \( r \). 
The circle touches the y-axis, so the distance from the center \( (h, k) \) to the y-axis must be equal to the radius \( r \). 
The distance from the point \( (h, k) \) to the y-axis is simply \( |h| \), so we have: \[ |h| = r. \] 
Step 2: Next, the circle touches the line \( x + y = 0 \), so the distance from the center \( (h, k) \) to this line must also be equal to the radius \( r \). The formula for the distance from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by: 
\[ {Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \] 
For the line \( x + y = 0 \), we have \( A = 1 \), \( B = 1 \), and \( C = 0 \). 
Thus, the distance from the center \( (h, k) \) to the line is: 
\[ \frac{|h + k|}{\sqrt{1^2 + 1^2}} = \frac{|h + k|}{\sqrt{2}}. \] 
Since this distance must equal the radius \( r \), we have: \[ \frac{|h + k|}{\sqrt{2}} = r. \] 
Step 3: Now, equating the two expressions for the radius \( r \), we get: \[ |h| = \frac{|h + k|}{\sqrt{2}}. \] 
Squaring both sides: 
\[ h^2 = \frac{(h + k)^2}{2}. \] 
Multiplying through by 2: \[ 2h^2 = (h + k)^2. \] 
Expanding the right-hand side: \[ 2h^2 = h^2 + 2hk + k^2. \] 
Simplifying: 
\[ h^2 = 2hk + k^2. \] Rearranging: \[ h^2 - k^2 = 2hk. \] 
Thus, the locus of the center of the circle is given by: \[ x^2 - y^2 = 2xy. \]