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 touches the x-axis, thus the radius $r = |a|$.
Step 2: Determine the intercept on the y-axis. The length of the intercept is $b$, which means $b = 2r$. Since it touches the x-axis at $a$, $b = 2|a|$.
Step 3: Calculate the coordinates of the center. Center $(h, k)$ is $(a, -a)$ because it lies below the x-axis.
Step 4: Substitute into the circle equation. $$(x - a)^2 + (y + a)^2 = a^2$$ Expanding and simplifying gives us the general form of the circle.
Step 5: Extract the coefficients and solve for the ordered pair. $$2a = \alpha, \quad b^2 = 4a^2 = \beta^2 + 4\gamma$$