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 $x-a x i s$

Accepted Answer

By pytogorus ${r}^{2}={a}^{2}+\frac{{b}^{2}}{4}={P}^{2}$

$r=\sqrt{\frac{4{a}^{2}+{b}^{2}}{4}}$

Equation of circle is $(x-\alpha {)}^{2}+(y-\beta {)}^{2}={r}^{2}$

$\begin{gathered} {x}^{2}+{y}^{2}-2ax-2py+{\alpha }^{2}+{p}^{2}-{r}^{2}=0 \ {x}^{2}+{y}^{2}-\alpha x+\beta y+r=0 \ -\alpha =-2a,\beta =-2p,r={a}^{2} \ \Rightarrow 2a=\alpha ,4{a}^{2}+{b}^{2}=4{p}^{2} \ {\alpha }^{2}+{b}^{2}=4{p}^{2} \ {\alpha }^{2}+{b}^{2}={\beta }^{2} \end{gathered}$

So, $(2a,{b}^{2})=(\alpha ,{\beta }^{2}-4r)$

Solutions · 1