If A=(0,-2) and B is any point on the circle | vedclass

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(B) The equation of the circle is $x^2+y^2-2x-2y+1=0$. Rewriting it in standard form: $(x-1)^2+(y-1)^2 = 1^2+1^2-1 = 1$. The center of the circle is $

Vedclass · Accepted Answer

$11+2\sqrt{10}$

Vedclass · Answer

(B) The equation of the circle is $x^2+y^2-2x-2y+1=0$. Rewriting it in standard form: $(x-1)^2+(y-1)^2 = 1^2+1^2-1 = 1$. The center of the circle is $C=(1,1)$ and the radius is $r=1$. The distance between point $A=(0,-2)$ and the center $C=(1,1)$ is $AC = \sqrt{(1-0)^2+(1-(-2))^2} = \sqrt{1^2+3^2} = \sqrt{10}$. The maximum distance $AB$ is given by $AC+r = \sqrt{10}+1$. Therefore,the maximum value of $(AB)^2$ is $(\sqrt{10}+1)^2 = 10+1+2\sqrt{10} = 11+2\sqrt{10}$.

If $A=(0,-2)$ and $B$ is any point on the circle $x^2+y^2-2x-2y+1=0$,then the maximum value of $(AB)^2$ is

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