Let the equation of the circle, which touches $x$-axis at the point $(a, 0), a \gt 0$ 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 $\left(2 a, b^2\right)$ is equal to

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Aicharya · Accepted Answer

By pythagoras $\mathrm{r}^2=\mathrm{a}^2+\frac{\mathrm{b}^2}{4}=\mathrm{P}^2$
$r=\sqrt{\frac{4 a^2+b^2}{4}}$
Equation of circle is $(x-\alpha)^2+(y-\beta)^2=r^2$
$x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0$
comparision $x^2+y^2-\alpha x+\beta y+r=0$
$\begin{array}{r}
-\alpha=-2 a, \beta=-2 p, r=a^2 \\ \Rightarrow 2 a=\alpha, 4 a^2+b^2=4 p^2 \\ \alpha^2+b^2=4 p^2 \\ \alpha^2+b^2=\beta^2
\end{array}$
So, $\left(2 \mathrm{a}, \mathrm{b}^2\right)=\left(\alpha, \beta^2-4 \mathrm{r}\right)$

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

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