Question

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

Answer 1

Answer:

(x+1)(x+2)+(y-1)(y+3)=0

or. x^2+3x+2+y^2+2y-3=0

or. (x+3/2)^2+(y+1)^2=1+9/4+1= 17/4

or. (x+3/2)^2 +(y+1)^2 = (√17/2)^2

Thus , coordinates of center (-3/2 , -1) and radius of circle is √17/2 units

Area of circle = π.r^2 = π.17/4= 22×17/4= 187/2= 93.5 units ^2. Answer.

Step-by-step explanation:

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Answer 2

Given: the circle$(x+1)(x+2)+(y-1)(y+3)=0$.

To find: the area of the circle.

Solution:

Know that,

$\[\begin{array}{l}(x + 1)(x + 2) + (y - 1)(y + 3) = 0\\ \Rightarrow {x^2} + 3x + 2 + {y^2} + 2y - 3 = 0\\ \Rightarrow {\left( {x + \frac{3}{2}} \right)^2} + {(y + 1)^2} = 1 + \frac{9}{4} + 1\end{array}\]$

$\[\begin{array}{l} \Rightarrow {\left( {x + \frac{3}{2}} \right)^2} + {(y + 1)^2} = \frac{{17}}{4}\\ \Rightarrow {\left( {x + \frac{3}{2}} \right)^2} + {(y + 1)^2} = {\left( {\frac{{\sqrt {17} }}{2}} \right)^2}\end{array}\]$

Therefore, coordinates of center$\[\left( { - \frac{3}{2},{\rm{ }} - 1} \right)\]$ and radius of circle is $\[\frac{{\sqrt {17} }}{2}\]$units.

Find the area of the circle.

Area of circle $\[ = \pi {r^2}\]$

                      $\[\begin{array}{l} = \frac{{22}}{7} \times \frac{{\sqrt {17} }}{2} \times \frac{{\sqrt {17} }}{2}\\ = \frac{{22}}{7} \times \frac{{17}}{4}\\ = \frac{{187}}{2}\\ = 93.5\,sq.units\end{array}\]$

Hence, the area of the circle is $93.5 sq. units$.