Find the equation of the circle with centre -1- 1- and radius displaystyle sqrt-2-

Question

Toppr Admin · Accepted Answer

Here it is given that centre -h-k-1-1- and radius r-x221A-2Therefore the equation of the circle is-x-x2212-1-2-y-x2212-1-2-x221A-2-2-x21D2-x2-x2212-2x-1-y2-x2212-2y-1-2-x21D2-x2-y2-x2212-2x-x2212-2y-0

Find the equation of the circle with centre $(1, 1)$ and radius $\sqrt{2}$

Here it is given that centre $(h, k) = (1, 1)$ and radius $r = \sqrt{2}$
Therefore the equation of the circle is
${(x - 1)}^{2} + {(y - 1)}^{2} = {(\sqrt{2})}^{2}$
$\Rightarrow x^{2} - 2 x + 1 + y^{2} - 2 y + 1 = 2$
$\Rightarrow x^{2} + y^{2} - 2 x - 2 y = 0$

Find the equation of the circle with centre (1,1) and radius √2

Here it is given that centre (h,k)=(1,1) and radius r=√2Therefore the equation of the circle is(x−1)2+(y−1)2=(√2)2⇒x2−2x+1+y2−2y+1=2⇒x2+y2−2x−2y=0

Find the equation of the circle with centre $(1, 1)$ and radius $\sqrt{2}$

Here it is given that centre $(h, k) = (1, 1)$ and radius $r = \sqrt{2}$
Therefore the equation of the circle is
${(x - 1)}^{2} + {(y - 1)}^{2} = {(\sqrt{2})}^{2}$
$\Rightarrow x^{2} - 2 x + 1 + y^{2} - 2 y + 1 = 2$
$\Rightarrow x^{2} + y^{2} - 2 x - 2 y = 0$