Question ID: #436
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:
- (1) $(\alpha, \beta^2 + 4\gamma)$
- (2) $(\gamma, \beta^2 – 4\alpha)$
- (3) $(\gamma, \beta^2 + 4\alpha)$
- (4) $(\alpha, \beta^2 – 4\gamma)$
Solution:
Since the circle touches the x-axis at $(a, 0)$ and lies below the x-axis, its center is at $(a, -r)$ where $r$ is the radius.
The equation of the circle is:
$$ (x-a)^2 + (y+r)^2 = r^2 $$
$$ x^2 – 2ax + a^2 + y^2 + 2ry + r^2 = r^2 $$
$$ x^2 + y^2 – 2ax + 2ry + a^2 = 0 $$
Comparing this with the given general equation $x^2 + y^2 – \alpha x + \beta y + \gamma = 0$:
1) $-\alpha = -2a \Rightarrow \alpha = 2a$
2) $\beta = 2r$
3) $\gamma = a^2$
We need to find an expression for $b^2$ (square of the y-intercept).
The length of the y-intercept for the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $2\sqrt{f^2 – c}$.
Here, $2f = \beta \Rightarrow f = \beta/2$ and $c = \gamma$.
$$ b = 2\sqrt{\left(\frac{\beta}{2}\right)^2 – \gamma} $$
$$ b^2 = 4\left(\frac{\beta^2}{4} – \gamma\right) $$
$$ b^2 = \beta^2 – 4\gamma $$
We need the ordered pair $(2a, b^2)$.
Substitute $\alpha = 2a$ and $b^2 = \beta^2 – 4\gamma$:
$$ (2a, b^2) = (\alpha, \beta^2 – 4\gamma) $$
Ans. (4)
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