Answered

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What is the area of the circle 4x^2 +4y^2 - 7x +6y -35 = 0

Sagot :

Havingtroublesidn
The equation of the circle is given by:
[tex]4 x^{2} +4 y^{2} -7x+6y-35=0 [/tex]

We could transform this to its general form [tex](x-h)^2+(y-k)^2=r^2[/tex] by manipulating the equation. (Mostly by completing a square.)

Rearranging the equation to easily see like terms, we get:
[tex]4x^2 -7x +4y^2 +6y -35=0[/tex]

Using the techniques of completing the square, we will have:
[tex](4x^2-7x+ \frac{49}{16}) - \frac{49}{16} +(4y^2+6y+\frac{3}{2}) -\frac{3}{2} -35=0[/tex]

Rearranging:
[tex](4x^2-7x+ \frac{49}{16})+(4y^2+6y+ \frac{9}{4}) = \frac{49}{16} + \frac{9}{4}+35[/tex]

Combining:
[tex](2x+ \frac{7}{4})^2 +(2x+ \frac{3}{2})^2= (\frac{ \sqrt{645} }{4} )^2[/tex]

Now, since we have the general equation and we know the value of r, which is [tex] \frac{ \sqrt{645} }{4} [/tex], we use the formula of getting the area of a circle in terms of r. 

Area= πr^2

Therefore the area is:
[tex]\frac{645 \pi }{16} [/tex]

Hope that helps.

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