Question

A circle touches both coordinate axes and its center lies on the line \( x - 2y - 3 = 0 \). What is its equation?

• A. \( x^2 + y^2 - 2x + 2y + 1 = 0 \)
• B. \( x^2 + y^2 + 2x - 2y + 1 = 0 \)
• C. \( x^2 + y^2 + 6x + 6y - 9 = 0 \)
• D. \( x^2 + y^2 - 6x - 6y + 9 = 0 \)

Hint:

Circle touching both axes has equal x and y center coordinates. Substitute into given line to solve.

Answer 1

The Correct Option is A

If a circle touches both coordinate axes, then its center is at \( (r, r) \) Now, center lies on line: \[ x - 2y - 3 = 0 \Rightarrow r - 2r - 3 = 0 \Rightarrow -r = 3 \Rightarrow r = -3 \] But radius can't be negative \Rightarrow Error in direction Take \( x = r,\ y = r \Rightarrow r - 2r - 3 = 0 \Rightarrow -r = 3 \Rightarrow r = -3 \) contradicts geometry. Try again: Let center be \( (a, a) \), substitute in line: \[ \begin{align} a - 2a - 3 = 0 \Rightarrow -a = 3 \Rightarrow a = -3 \Rightarrow \text{invalid}
\Rightarrow \text{Try positive radius: } (a, a) = (1, 1)
\Rightarrow x - 2y - 3 = 1 - 2(1) - 3 = -4 \Rightarrow no Eventually, point \( (1, -1) \) satisfies: \[ x - 2y - 3 = 0 \Rightarrow 1 + 2 - 3 = 0 \] So center \( (1, -1) \), radius \( = 1 \) Equation: \[ (x - 1)^2 + (y + 1)^2 = 1 \Rightarrow x^2 + y^2 - 2x + 2y + 1 = 0 \]