Let the equation of the circle, which touches x-axis at the point (a, 0), a0 and cuts off…

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 $y-a x i s$ 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

Accepted Answer

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, $\left(2 a, b^2\right)=\left(\alpha, \beta^2-4 r\right)$

Solutions · 1