JEE MAIN - Mathematics (2025 - 28th January Morning Shift - No. 17)
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 $y-a x i s$ be $x^2+y^2-\alpha x+\beta y+\gamma=0$. If the circle lies below $x-a x i s$, then the ordered pair $\left(2 a, b^2\right)$ is equal to
$\left(\alpha, \beta^2+4 \gamma\right)$
$\left(\alpha, \beta^2-4 \gamma\right)$
$\left(\gamma, \beta^2-4 \alpha\right)$
$\left(\gamma, \beta^2+4 \alpha\right)$
Explanation
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By pytogorus $\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2$
$$r=\sqrt{\frac{4 a^2+b^2}{4}}$$
Equation of circle is $(x-\alpha)^2+(y-\beta)^2=r^2$
$$\begin{aligned} & x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0 \\ & \text { comparision } x^2+y^2-\alpha x+\beta y+r=0 \\ & -\alpha=-2 a, \beta=-2 p, r=a^2 \\ & \Rightarrow 2 a=\alpha, 4 a^2+b^2=4 p^2 \\ & \alpha^2+b^2=4 p^2 \\ & \alpha^2+b^2=\beta^2 \end{aligned}$$
So, $\left(2 a, b^2\right)=\left(\alpha, \beta^2-4 r\right)$
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