Let a circle of radius 4 pass through the origin O , the points A(-√3a,0) and B(0,-√2b), w…

Question

Let a circle of radius 4 pass through the origin O , the points $A(-\sqrt{3}a,0)$ and $B(0,-\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab
e 0$. Then the locus of the centroid of $	riangleOAB$ is a circle of radius

Accepted Answer

Given a circle passes through $O(0,0),A(-\sqrt{3}a,0)$, and $B(0,-\sqrt{2}b)$. Since $A$ lies on the x -axis and $B$ lies on the y -axis, $\angleAOB={90}^{\circ }$. Thus $AB$ is the diameter of the circle.

Diameter $d=2	imes$ radius $=2(4)=8$.

Using distance formula for $AB$ :

$\begin{gathered} \sqrt{(-\sqrt{3}a-0{)}^{2}+(0-(-\sqrt{2}b){)}^{2}}=8 \ 3{a}^{2}+2{b}^{2}=64\, ------(1) \end{gathered}$

Let the centroid of $	riangleOAB$ be $G(h,k)$.

$\begin{gathered} h=\frac{0-\sqrt{3}a+0}{3}\implies a=\frac{-3h}{\sqrt{3}}=-\sqrt{3}h \ k=\frac{0+0-\sqrt{2}b}{3}\implies b=\frac{-3k}{\sqrt{2}} \end{gathered}$

Substitute $a$ and $b$ into (Eq. 1):

$\begin{gathered} 3(-\sqrt{3}h{)}^{2}+2{(\frac{-3k}{\sqrt{2}})}^{2}=64 \ 3(3{h}^{2})+2(\frac{9{k}^{2}}{2})=64 \ 9{h}^{2}+9{k}^{2}=64 \ {h}^{2}+{k}^{2}=\frac{64}{9} \end{gathered}$

The locus is a circle ${x}^{2}+{y}^{2}={(\frac{8}{3})}^{2}$.

Radius $=\frac{8}{3}$

Solutions · 1