maths q (1 Viewer)

blacktown_ · Dec 3, 2012

find the length of the diameter of the circle x^2+8x+y^2-12y+3=0

please show all working out!!

SpiralFlex · Dec 3, 2012

$x^2+8x+16-16+y^2-12y+36-36+3=0$

$(x+4)^2+(y-6)^2=49$

We have a circle in the form of,

$(x-h)^2+(y-k)^2=r^2$

$r=7, \therefore d=14$

A more mental approach is,

Using the general form of a circle,

$x^2 + y^2 + 2gx + 2fy + c = 0$

Where the diameter is,

$d=2\sqrt{g^2+f^2-c}$

$=2\sqrt{4^2+(-6)^2-3}$

$=14$

blacktown_ · Dec 3, 2012

omg, thanks so much SpiralFlex

maths q (1 Viewer)

blacktown_

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SpiralFlex

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blacktown_

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