If the four distinct points $(4,6),(-1,5),(0,0)$ and $(\mathrm{k}, 3 \mathrm{k})$ lie on a circle of radius r , then $10 \mathrm{k}+\mathrm{r}^2$ is equal to

Question

Aicharya · Accepted Answer

$\mathrm{m}_1 \mathrm{~m}_2=-1$ so right angle equation circle is
$\begin{aligned}
& (x-4)(x-0)+(y-6)(y-0)=0 \\ & x^2+y^2-4 x-6 y=0
\end{aligned}$
$(\mathrm{k}, 3 \mathrm{k})$ lies on it so
$\begin{aligned}
& \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-18 \mathrm{k}=0 \\ & 10 \mathrm{k}^2-22 \mathrm{k}=0
\end{aligned}$
$\mathrm{k}=0, \frac{11}{5}$
$\mathrm{k}=0$ is not possible so $\mathrm{k}=\frac{11}{5}$
also $\mathrm{r}=\sqrt{4+9}=\sqrt{13}$
so $10 \mathrm{k}+\mathrm{r}^2=10 \cdot \frac{11}{5}+(\sqrt{13})^2=35$

If the four distinct points $(4,6),(-1,5),(0,0)$ and $(\mathrm{k}, 3 \mathrm{k})$ lie on a circle of radius…

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