Q90 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics

If the polar of a point $P$ with respect to a circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is $x+2 y=4 r$, then the point $P$ is

  • A
    $(r, 2 r)$
  • B
    $(2 r, r)$
  • Correct answer
    C
    $(2 r, 3 r)$
  • D
    $(-r, 4 r)$

Solution

Let $P$ be $\left(x_1, y_1\right)$ Now, equation of circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is $\begin{aligned} (x-r)^2+(y-r)^2 & =r^2 \\ \Rightarrow \quad x^2+y^2-2 x r-2 y r-r^2 & =0 \end{aligned}$ Now, polar of $P$ with respect to the circle $\begin{aligned} & x^2+y^2-2 x r-2 y r-r^2=0 \text { is } \\ & \quad x x_1+y y_1-\left(x+x_1\right) r-\left(y+y_1\right) r-r^2=0 \\ & \Rightarrow \quad\left(x_1-r\right) x+\left(y_1-r\right) y=\left(x_1+y_1-r\right) r \end{aligned}$ But it is given that polar of $P$ with respect to above circle is $x+2 y=4 r$ $\begin{aligned} & \therefore \quad \frac{x_1-r}{1}=\frac{y_1-r}{2}=\frac{x_1+y_1-r}{4} \\ & \Rightarrow \quad x_1-r=\frac{x_1+y_1-r}{4} \text { and } x_1-r=\frac{y_1-r}{2} \\ & \Rightarrow \quad 4 x_1-4 r=x_1+y_1-r \text { and } 2 x_1-2 r=y_1-r \\ & \Rightarrow \quad 3 x_1-3 r=y_1 \text { and } 2 x_1-r=y_1 \\ & \therefore \quad 3 x_1-3 r=2 x_1-r \\ & \Rightarrow \quad \quad x_1=2 r \text { and } y_1=2 x_1-r=4 r-r=3 r \\ & \therefore \quad P \text { is }(2 r, 3 r) \end{aligned}$

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

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