Mathematics - Circle Question with Solution | TestHub

Slope of $\mathrm{PC}=\frac{-3-2}{2-7}=1$

if $\tan \theta =1\because \theta =45^{\circ }$

Equation of PA is $\frac{x-7}{1/\sqrt{2}}=\frac{y-2}{1/\sqrt{2}}=r$

$\therefore \left(7+\frac{r}{\sqrt{2}},2+\frac{r}{\sqrt{2}}\right)$ lie on circle, then, ${\left(7+\frac{r}{\sqrt{2}}\right)}^2+{\left(2+\frac{r}{\sqrt{2}}\right)}^2-4\left(7+\frac{r}{\sqrt{2}}\right)+t\left(2+\frac{r}{\sqrt{2}}\right)-3=0$

$\Rightarrow {\mathrm{r}}^2+10\sqrt{2}\mathrm{r}+34=0$ $\therefore \mathrm{r}=-5\sqrt{2}\pm 4$

$$\begin{array}{rlr}& \therefore \text{ Point }\left(7+\frac{-5\sqrt{2}\pm 4}{\sqrt{2}},2+\frac{-5\sqrt{2}\pm 4}{\sqrt{2}}\right)& \Rightarrow (2\pm 2\sqrt{2},-3\pm 2\sqrt{2})\end{array}$$

taking + ve sign for $\mathrm{A},-$ ve sign for B .