Suppose two tangents PA and PB are drawn to the circle centered at $\mathrm{C}(1,2)$ from the point…

Suppose two tangents PA and PB are drawn to the circle centered at $\mathrm{C}(1,2)$ from the point $\mathrm{P}(16,7)$. If the area of the quadrilateral PACB is 75 square units, then the radius of the circle is
  1. 5
  2. 25
  3. 225
  4. $\sqrt{5}$

Solution

$\mathrm{PC}=\sqrt{(16-1)^2+(7-2)^2}=\sqrt{250}$ Area of PACB $ =\mathrm{AP} \times \mathrm{AC}=\mathrm{xr}=75 \Rightarrow \mathrm{x}=\frac{75}{\mathrm{r}} $ In triangle APC $ \begin{aligned} & \mathrm{x}^2+\mathrm{r}^2=\mathrm{PC}^2 \Rightarrow\left(\frac{75}{\mathrm{r}}\right)^2+\mathrm{r}^2=250 \\ & \mathrm{r}^4-250 \mathrm{r}^2+5625=0 \\ & \mathrm{r}^4-225 \mathrm{r}^2-25 \mathrm{r}^2+5625=0 \\ & \Rightarrow\left(\mathrm{r}^2-225\right)\left(\mathrm{r}^2-25\right)=0 \\ & \Rightarrow \mathrm{r}^2=225 \Rightarrow \mathrm{x}=15 \\ & \text { or } \mathrm{r}^2-25=0 \Rightarrow \mathrm{r}=5 \end{aligned} $

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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