How do you find the center and radius for $X^{2} + Y^{2} = 13$ $X^2 + Y^2 = 13$ ?

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Answer 1 · 2016-01-07T23:37:39

centre = (0 , 0 ) and r = $\sqrt{13}$ $sqrt13$

require to compare this equation with the standard form of the equation of a circle.

${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ $(x - a )^2 + (y - b )^2 = r^2$

where (a , b ) represents the centre of the circle and r the radius

$X^{2} + Y^{2} = {(X - 0)}^{2} + {(Y - 0)}^{2}$ $X^2 + Y^2 = (X- 0 )^2 + (Y -0 )^2$

$r^{2} = 13 \Rightarrow r = \sqrt{13}$ $r^2 = 13 rArr r = sqrt13$

hence centre = (0 ,0 ) and radius = $\sqrt{13}$ $sqrt13$

How do you find the center and radius for X^2 + Y^2 = 13X2+Y2=13X^2 + Y^2 = 13?