Question

Let the equation of the circle, which touches x-axis at the point \( (a, 0) \) and cuts off an intercept of length \( b \) on y-axis be \( x^2 + y^2 - cx + dy + e = 0 \). If the circle lies below x-axis, then the ordered pair \( (2a, b^2) \) is equal to:

• A. \( (y, \beta^2 - 4\alpha) \)
• B. \( (\alpha, \beta^2 - 4\gamma) \)
• C. \( (y, \beta^2 + 4\alpha) \)
• D. \( (\alpha, \beta^2 + 4\gamma) \)

Hint:

Always visualize the geometry of the circle in relation to coordinate axes to better understand its equation and properties.

Answer 1

The Correct Option is D

Step 1: Define the geometry of the circle. The circle is tangent to the x-axis, so its radius is \( r = |a| \).
Step 2: Determine the y-axis intercept. The length of the intercept is \( b \), where \( b = 2r \). Since the circle is tangent to the x-axis at \( a \), the intercept length is \( b = 2|a| \). 
Step 3: Calculate the center coordinates. The center \( (h, k) \) is \( (a, -a) \) as it is situated below the x-axis. 
Step 4: Substitute into the circle equation. The equation becomes \( (x - a)^2 + (y + a)^2 = a^2 \). Expansion and simplification yield the general form of the circle. Step 5: Extract coefficients and solve for the ordered pair. From the general form, we have \( 2a = \alpha \) and \( b^2 = 4a^2 = \beta^2 + 4\gamma \).