The equation of the circle touching the lines | vedclass

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Question

(A) The given equation $|x-2|+|y-3|=4$ represents a square with vertices at $(2+4, 3) = (6, 3)$,$(2-4, 3) = (-2, 3)$,$(2, 3+4) = (2, 7)$,and $(2, 3-4)

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$x^2+y^2-4x-6y+5=0$

Vedclass · Answer

(A) The given equation $|x-2|+|y-3|=4$ represents a square with vertices at $(2+4, 3) = (6, 3)$,$(2-4, 3) = (-2, 3)$,$(2, 3+4) = (2, 7)$,and $(2, 3-4) = (2, -1)$.This square is centered at $(2, 3)$ and has a side length of $4\sqrt{2}$.The circle touching these lines is the incircle of this square.The center of the circle is the same as the center of the square,which is $(h, k) = (2, 3)$.The radius $r$ of the incircle is half the distance between parallel sides,or the perpendicular distance from the center $(2, 3)$ to any of the lines,such as $(x-2)+(y-3)=4 \implies x+y-9=0$.The distance $r = \frac{|2+3-9|}{\sqrt{1^2+1^2}} = \frac{|-4|}{\sqrt{2}} = 2\sqrt{2}$.The equation of the circle is $(x-h)^2+(y-k)^2=r^2$.Substituting the values: $(x-2)^2+(y-3)^2=(2\sqrt{2})^2$.$x^2-4x+4+y^2-6y+9=8$.$x^2+y^2-4x-6y+5=0$.

The equation of the circle touching the lines $|x-2|+|y-3|=4$ is

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