Question

A big confusion, is it
$1)r = \sqrt{g^{2} + f^{2} + c }$
OR
$2)r = \sqrt{ {g}^{2} + {f}^{2} - c}$

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Answer 1

r= √g^2 + f^2 - c

let's derive

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

Centre we know is

( -g, - f)

Differentiate it

2x + 2 y y' + 2g + 2f y' = 0

x + g + y'(y + f) = 0

y' = -( x+ g) /( y+f)

y = -( x+ g)/ ( y+ f) x + c

Well its tedious

lets do by another

As x= -g + r cos a

y= - f + r sin a

its parametric coordinates which is obvious as any coordinate from centre

and then take along x axis and y axis its component that is r cos and r sin and its x coirdinate would be centre x or y + component

So substitute in eq

( - g + r cos ) ^2 + ( -f+ r sin)^2 + 2 g( -g + r cos) + 2 f( - f+ rsin) + c= 0

g^2 + r^2 cos^2 - 2g r cos + f^2 + r^2 sin^2 -2fr sin - 2g^2 + 2gr cos - 2 f^2 + 2fr sin + c= 0

r^2( cos^2 + sin^2) - g^2 - f^2 + c=0

As sin^2 + cos^2 = 1

r^2 - g^2 - f^2 + c= 0

r^2 = g^2 + f^2 - c

r= √g^2 + f^2 - c

✌✌✌Dr.Dhruv✌✌✌✌