# What is the area of the circle x^2 + y^2 – 8x – 6y = 0?

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We can write the given equation of the circle, x^2 + y^2 – 8x – 6y = 0 as

x^2 + y^2 – 8x – 6y = 0

=> x^2 – 8x + y^2 – 6y = 0

=> x^2 – 8x + 16 + y^2 – 6y +9 = 16 + 9

=> (x – 4) ^2 + (y – 3) ^2 = 25 = 5^2

Therefore the radius of the circle is 5.

The area of a circle is given by pi*r^2 = pi*5^2 = 25*pi.

**The required area of the circle represented by the equation x^2 + y^2 – 8x – 6y = 0 is 25*pi.**

The area A of a circle is given by A = pi*r^2, where r is the radius of the circle.

The given equation of the circle is x^2 + y^2 – 8x – 6y = 0. We recast this equation into centre radius form of (x-h)^2+(y-k)^2= r^2.........(1), where (h,k) is the coordinates of the centre and r is the radius:

x^2+y^2-8x-6y = 0.

=> (x^2-8x) +y^2-6y) = 0.

We add 4^2+3^2 to both sides as below:

(x^2-8x+4^2)+(y^2-6x+3^2) = 4^2+3^2.

(x-4)^2+(y-3)^2 = 5^2....(2).

Since (1) and (2) represent the same circle, (h,k) = (4,3) and r = 5.

So the given equation has the centre at (4,3) and its radius is 5.

So the area of the circle is A = pir^2 = Pi*5^2 = 25pi = 78.54 unit^2 , nearly.