If the centre of a circle is (2, 3) and a tan | vedclass

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Question

(A) The radius $r$ of the circle is the perpendicular distance from the centre $(2, 3)$ to the tangent line $x + y - 1 = 0$.Using the formula for perp

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$(x - 2)^2 + (y - 3)^2 = 8$

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(A) The radius $r$ of the circle is the perpendicular distance from the centre $(2, 3)$ to the tangent line $x + y - 1 = 0$.Using the formula for perpendicular distance $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we get:$r = \frac{|1(2) + 1(3) - 1|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 1|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The square of the radius is $r^2 = (2\sqrt{2})^2 = 8$.The equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.Substituting $(h, k) = (2, 3)$ and $r^2 = 8$,we get $(x - 2)^2 + (y - 3)^2 = 8$.

If the centre of a circle is $(2, 3)$ and a tangent is $x + y = 1$,then the equation of this circle is

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