Let the centre of the circle $\mathrm{S}=0$ lie on the line $x+y-5=0$ and also lie in the first quadrant. If this circle touches both the lines $x-2=0$ and $y-5=0$, then the area of the circle is

• Correct answer
  A

  $\pi$ sq. units
• B

  $2 \pi$ sq. units
• C

  $4 \pi$ sq. units
• D

  $\frac{1}{4} \pi$ sq. units

Solution

Given a circle $5=0$ whose centre lies on the line $\mathrm{x}+\mathrm{y}-5=0$ Let the centre of circle -be,$(h, k)$

So, $(h, 5)$ and $(2, k)$ also Satisfy the equation of circle. Equation of circle is $(x-4)^2+(y-k)^2=r^2$. Satisfy the points (h, 5) and $(2, \mathrm{k})$ $$ \begin{aligned} & (\mathrm{h}-\mathrm{h})^2+(5-\mathrm{k})^2=\mathrm{r}^2 \\ & 5-\mathrm{k}=\mathrm{r} \ldots \ldots\left(\mathrm{a}^{\prime}\right) \\ & \left.(2=\mathrm{h})^2+\mathrm{k}-\mathrm{k}\right)^2=\mathrm{r}^2 \\ & 2=\mathrm{h}=\mathrm{r} \ldots \ldots . .\left(\mathrm{b}^{\prime}\right) \\ & \text { Here, }\left(\mathrm{a}^{\prime}\right)=\left(\mathrm{b}^{\prime}\right) \\ & 5-\mathrm{k}=2-\mathrm{h} \end{aligned} $$
Add (i) \& (ii) $$ \begin{aligned} & \mathrm{h}+\mathrm{k}=5 \\ & \frac{\mathrm{h}-\mathrm{k}=-3}{2 \mathrm{~h}=2} \\ & \mathrm{~h}=1 \end{aligned} $$ from (i) $$ \mathrm{h}+\mathrm{k}=5 $$ $$ \mathrm{k}=4 $$ Then, $(\mathrm{h}, \mathrm{k}) \rightarrow(1,4)$ from $\left(a^{\prime}\right)$ $$ \Rightarrow \mathrm{r}=5-4=1 $$ Required area of circle $=\pi \mathrm{r}^2=\pi(1)^2=\pi$ So, option (a) is correct

Hence, the correct answer is option: 'A'.