What is the area of the circle whose equation is x^2 + y^2 - 2x + 4y = 32

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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Given the equation of the circle is:

x^2 + y^2 -2x + 4y = 32

We need to find the area of the circle.

To find the area, we need to determine the radius of the circle (r).

We can determine r by rewriting the circle equation into the standard form.

 (x-a)^2 + (y-b)^2 = r^2 where r is the radius.

Let us rewrite by completing the square.

==> x^2 + y^2 - 2x + 4y = 32

==> x^2 - 2x + y^2 + 4y = 32

==> x^2 - 2x + 1 -1 + y^2 + 4y + 4 - 4 = 32

==> (x-1)^2 -1 + ( y+2)^2 - 4 = 32

==> (x-1)^2 + (y+2)^2 = 32 + 5

==> (x-1)^2 + (y+2)^2 = 37

Then, we conclude that r^2 = 37.

==> r= sqrt(37).

Now we will determine the area.

==> The area (a )= r^2 * pi = 37*pi = 116.24 square units (approx.)

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

 What is the area of the circle whose equation is x^2 + y^2 - 2x + 4y = 32.

The are of the circle is pi*r^2, where r is the radius.

The radius r of the circle whose equation is x^2+y^2+2gx+2fy+c = 0 is given by:

r = sqrt(g^2+g^2-c)....(1).

So we write the given equation x^2 + y^2 - 2x + 4y = 32 in the standard form x^2+y^2+2gx+2fy+c = 0 as below:

x^2 + y^2 - 2x + 4y - 32 = 0.

S0 comparing  2g = -2, 2f = 4 and c= -32.

Therefore g= -2/2 = -1 and f = 4/2 = 2..

Therefore = sqrt(g^2+f^2-c) = sqrt{(-1)^2+2^2-(-32} = sqrt(1+4+32} = sqrt37.

Therefore the radius of the given circle is sqrt37.

So the area of the circle is p*r^2 = pi*(sqrt37)^2= 37pi = 116.24 sq units.

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