Find the area of the circle given by the equa | vedclass

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Question

(A) The given equation of the circle is $(x+1)(x+2)+(y-1)(y+3)=0$. Expanding the terms,we get $x^2+3x+2+y^2+2y-3=0$,which simplifies to $x^2+y^2+3x+2y

Vedclass · Accepted Answer

$\frac{17 \pi}{4}$

Vedclass · Answer

(A) The given equation of the circle is $(x+1)(x+2)+(y-1)(y+3)=0$. Expanding the terms,we get $x^2+3x+2+y^2+2y-3=0$,which simplifies to $x^2+y^2+3x+2y-1=0$. Comparing this with the standard form $x^2+y^2+2gx+2fy+c=0$,we have $2g=3 \implies g=\frac{3}{2}$,$2f=2 \implies f=1$,and $c=-1$. The radius $r$ is given by $r=\sqrt{g^2+f^2-c} = \sqrt{(\frac{3}{2})^2 + (1)^2 - (-1)} = \sqrt{\frac{9}{4}+1+1} = \sqrt{\frac{9+8}{4}} = \sqrt{\frac{17}{4}}$. The area of the circle is $\pi r^2 = \pi (\frac{17}{4}) = \frac{17 \pi}{4}$. Thus,option $A$ is correct.

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

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