From a point $P(0, b)$ two tangents are drawn to the circle $x^2+y^2=16$ and these two tangents intersect…

1. $x^2+y^2=16 \sqrt{2}$
2. $x^2+y^2=64$
3. $x^2+y^2=32$
4. $x^2+y^2=4 \sqrt{2}$

Solution

Equation of pair of tangents from point $(0, b)$ drawn to the circle $x^2+y^2=16$ is
For point $A$ and $B$, put $y=0$, then we are getting $ \begin{aligned} & \left(x^2-16\right)\left(b^2-16\right)=16^2 \\ \Rightarrow \quad & x^2=\frac{16 b^2}{b^2-16} \Rightarrow x= \pm \frac{4 b}{\sqrt{b^2-16}} \end{aligned} $ So, $x$-coordinate of A and B is $\pm \frac{4 b}{\sqrt{b^2-16}}$ Now, area of $\triangle P A B=\Delta=\frac{1}{2} b\left(\frac{8 b}{\sqrt{b^2-16}}\right)=\frac{4 b^2}{\sqrt{b^2-16}}$ Now, for minimum area $\frac{d \Delta}{d b}=0$ $ \begin{aligned} \Rightarrow & & \sqrt{b^2-16}(8 b) & =4 b^2 \frac{b}{\sqrt{b^2-16}} \\ & \Rightarrow & 8\left(b^2-16\right) & =4 b^2 \\ & \Rightarrow & b^2 & =32 \\ & \Rightarrow & b & = \pm 4 \sqrt{2} \end{aligned} $ So, $x$ - coordinate of $A$ and $B$ is $\pm \frac{16 \sqrt{2}}{4}= \pm 4 \sqrt{2}$ So, $P(0, \pm 4 \sqrt{2}), A(4 \sqrt{2}, 0)$ and $B(-4 \sqrt{2}, 0)$ $\because \triangle P A B$ is a right angled triangle, so equation of circumcircle of $\triangle P A B$ is $x^2+y^2=32$

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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