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(A) Let $r$ be the radius of the circle. The center of the circle is $C(2,4)$.The perpendicular distance $d$ from the center $(2,4)$ to the line $x+y+

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$\sqrt{41}$

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(A) Let $r$ be the radius of the circle. The center of the circle is $C(2,4)$.The perpendicular distance $d$ from the center $(2,4)$ to the line $x+y+2=0$ is given by:$d = \frac{|2+4+2|}{\sqrt{1^2+1^2}} = \frac{8}{\sqrt{2}} = 4\sqrt{2}$.The chord length is $6$,so the distance from the center to the chord bisects the chord into two segments of length $3$. Let $A$ be the foot of the perpendicular from $C$ to the chord and $B$ be a point on the circle where the chord meets it.In the right-angled triangle $	riangle CAB$,by the Pythagorean theorem:$r^2 = (AC)^2 + (AB)^2$$r^2 = (4\sqrt{2})^2 + (3)^2$$r^2 = 32 + 9 = 41$$r = \sqrt{41}$.

$A$ circle with center at $(2,4)$ is such that the line $x+y+2=0$ cuts a chord of length $6$. The radius of the circle is

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