Circles
Circle
nta_pyq_2025_apr
Grade 11
Question:
Let the equation of the circle, which touches $x$-axis at the point $(a, 0)$, $a > 0$ and cuts off an intercept of length $b$ on $y$-axis be $x^2 + y^2 - \alpha x - \beta y + \gamma = 0$. If the circle lies below $x$-axis, then the ordered pair $(2a, b^2)$ is equal to
$(\gamma,\ \beta^2 - 4\alpha)$
$(\alpha,\ \beta^2 + 4\gamma)$
$(\gamma,\ \beta^2 + 4\alpha)$
$(\alpha,\ \beta^2 - 4\gamma)$
Step-by-Step Solution
Key Concept: A circle lying below the x-axis and touching it at $(a,0)$ has centre $(a,-r)$; expand the equation, match coefficients, and use the y-intercept chord-length formula $b^2 = \beta^2 - 4\gamma$.
Since the circle touches the x-axis at $(a,0)$ from below, its centre is $(a,-r)$ with radius $r > 0$. Expanding $(x-a)^2+(y+r)^2=r^2$: $$x^2+y^2-2ax+2ry+a^2=0.$$ Comparing with $x^2+y^2-\alpha x-\beta y+\gamma=0$: $\alpha=2a$, $\beta=-2r$, $\gamma=a^2$. Setting $x=0$: $y^2-\beta y+\gamma=0$, so $b^2=(\text{chord length})^2=\beta^2-4\gamma=4r^2-4a^2$. Hence $(2a,b^2)=(\alpha,\beta^2-4\gamma)$.
Correct Answer: 4