Find the centre and radius of the circle $x^{2} + y^{2} - 2 x - 2 y = 0$

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Solution

Given: $x^{2} + y^{2} - 2 x - 2 y = 0$

$\Rightarrow x^{2} - 2 x + 1 - 1 + y^{2} - 2 y + 1 - 1 = 0$

$\Rightarrow {(x - 1)}^{2} + {(y - 1)}^{2} - 2 = 0$

$\Rightarrow {(x - 1)}^{2} + {(y - 1)}^{2} = 2$

which is comparable with ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$

Here $h = 1, k = 1$ and $r = \sqrt{2}$

Hence, the given circle has centre $(1, 1)$ and its radius $= \sqrt{2}$

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Find the centre and radius of the circle x2+y2−2x−2y=0

Given: x2+y2−2x−2y=0⇒x2−2x+1−1+y2−2y+1−1=0⇒(x−1)2+(y−1)2−2=0⇒(x−1)2+(y−1)2=2which is comparable with (x−h)2+(y−k)2=r2Here h=1,k=1 and r=√2Hence, the given circle has centre (1,1) and its radius=√2

Find the centre and radius of the circle $x^{2} + y^{2} - 2 x - 2 y = 0$