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:
Show Hint
- \( (y, \beta^2 - 4\alpha) \)
- \( (\alpha, \beta^2 - 4\gamma) \)
- \( (y, \beta^2 + 4\alpha) \)
- \( (\alpha, \beta^2 + 4\gamma) \)
The Correct Option is D
Solution and Explanation
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 \( (\alpha, \beta^2+4\gamma) \).
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