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

The equation of the circle can be rewritten as (x - a)² + (y - r)² = r², where the circle touches the x-axis at the point (\(a, 0\)), meaning its radius \(r\) is such that the center of the circle is \( (a, -r) \). Thus, the circle has the form:

x² + y² - 2ax + 2ry + e = 0.

Since it touches the x-axis at \( (a, 0) \), the distance from \((a, -r)\) to the x-axis is \(r\), confirming \(b = 2r\), as it cuts an intercept \(b\) on the y-axis. Solving for the center's y-coordinate from the intercept, we have \( (0, b/2)\) implies:

\((0^2+(b/2)^2+r^2=b^2)\).

Simplifying:

• \(b^2/4 + r^2 = b^2\).
• \(r^2 = 3b^2/4\).

Additionally, the center's equation gives \(d = 2r = b\). Comparing the circle's form with:

• \(c = 2a\).
• \(d = -2r = -b\).

The conditions given in the options align \( (2a, b^2) \) with \((\alpha,\beta^2+4\gamma)\). Therefore, the matching conditions confirm:

• \(\alpha=2a\).
• \(\beta^2=b^2\).
• \(4\gamma=4r^2=3b^2\).

Thus, the correct option is \( (\alpha, \beta^2+4\gamma) \).