Let the equation of the circle, which touches the x-axis at the point (a, 0), a 0, and cuts off an intercept…

Question

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

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

Accepted Answer

B. $$ (\alpha, \beta^2 - 4\gamma) $$

Answer

A. $$ (\gamma, \beta^2 - 4\alpha) $$

Answer

C. $$ (\gamma, \beta^2 + 4\alpha) $$

Answer

D. $$ (\alpha, \beta^2 + 4\gamma) $$

Answer & step-by-step solution

Practice more Circle Equation & Properties questions

Related questions