Find the area of the circle $$ (x+1)(x+2)+(y-1)(y+3)=0 $$

Find the area of the circle $$ (x+1)(x+2)+(y-1)(y+3)=0 $$
  1. $\frac{17 \pi}{4}$
  2. $\frac{17 \pi}{2}$
  3. $\frac{2 \pi}{17}$
  4. $\frac{\pi}{3}$

Solution

Given circle is, $ \begin{aligned} &(x+1)(x+2)+(y-1)(y+3)=0 \\ & x^2+y^2+3 x+2 y-1=0 \\ & 2 g=3 \\ & g=\frac{3}{2} \\ & 2 f=2 \\ & f=1 \\ & c=-1 \\ & r=\sqrt{g^2+f^2-c}=\sqrt{\frac{9}{4}+1+1} \\ & r=\sqrt{\frac{17}{4}} \end{aligned} $ Hence, option (1) is correct

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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