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

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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…

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Given: The circle touches the x-axis at $(a,0)$ and lies below the x-axis. Its equation is $x^2 + y^2 - cx + dy + e = 0$ . It cuts an intercept of length $b$ on the y-axis.

Find: The ordered pair $(2a,b^2)$ in terms of the coefficients.

Since the circle touches the x-axis at $(a,0)$ and lies below the x-axis, its center is $(a,-r)$ for some radius $r$ . Hence its standard form is

(x-a)^2 + (y+r)^2 = r^2

Expanding,

x^2 + y^2 - 2ax + 2ry + a^2 = 0

Comparing with

x^2 + y^2 - cx + dy + e = 0

we get

c = 2a, \qquad d = 2r, \qquad e = a^2

To find the intercept cut on the y-axis, put $x=0$ in the circle equation:

y^2 + dy + e = 0

The two intersection points on the y-axis differ by length

b = \sqrt{d^2 - 4e}

Therefore,

b^2 = d^2 - 4e

Also, from the coefficient comparison above,

2a = c

Thus the ordered pair becomes

(2a,b^2) = (c, d^2 - 4e)

The solution explicitly marks Option D as correct, whose content is $(\alpha, \beta^2 + 4\gamma)$ . Preserving the listed options labels verbatim, the correct option is D.

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