Question

Let O be the centre of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$where $\mathrm{r}>\sqrt{\frac{5}{2}}$.

Answer 1

Finding the value of ‘r’:

Given Equation of Circle is,

${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2,r>\frac{\sqrt{5}}{2}$

‘O’ is the Centre of a circle.

$\mathrm{PQ}$ is the chord of this circle and the Equation of $\mathrm{PQ}$ is

$\mathrm{PQ}=2\mathrm{x}+4\mathrm{y}=5$

$\mathrm{C}$ is the Centre of a Circumcircle of triangle $\mathrm{OPQ}$ which lies on a line $\mathrm{x}+2\mathrm{y}=4$

Now,

$\mathrm{PQ}:\mathrm{hx}+\mathrm{ky}={\mathrm{x}}^2$

$$\begin{gathered} \Rightarrow 2\mathrm{x}+4\mathrm{y}=5 \\ \Rightarrow \mathrm{h}=2,\mathrm{k}=4\text{,}{\mathrm{r}}^2=5. \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{h}}{2}=\frac{\mathrm{k}}{4}=\frac{r^2}{5} \\ \mathrm{h}=\frac{2r^2}{5},\mathrm{k}=\frac{4r^2}{5} \end{gathered}$$

Therefore,

$\mathrm{C}\equiv \left[\frac{{\mathrm{r}}^2}{5},\frac{2r^2}{5}\right]$

This $\mathrm{C}$ lies on the line $\mathrm{x}+2\mathrm{y}=4$, therefore,

$\frac{r^2}{5}+2\left[2\frac{r^2}{5}\right]=4$

${\text{⇒ r}}^2\left(\frac{1}{5}+\frac{4}{5}\right)=4$

$\Rightarrow r^2\left(\frac{5}{5}\right)=4$

$$\begin{gathered} \Rightarrow r^2=4 \\ \Rightarrow r=2 \end{gathered}$$

Therefore the radius of a Circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$ is equal to $\mathbf{2}$.