The equation of circle with centre (1, 2) and | vedclass

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Question

(b) Radius of circle = perpendicular distance of tangent from the centre of circle

$ \Rightarrow $ $r = \frac{{1 + 2 - 5}}{{\sqrt {1 + 1} }} = \sqr

Vedclass · Accepted Answer

${x^2} + {y^2} - 2x - 4y + 3 = 0$

Vedclass · Answer

(b) Radius of circle = perpendicular distance of tangent from the centre of circle

$ \Rightarrow $ $r = \frac{{1 + 2 - 5}}{{\sqrt {1 + 1} }} = \sqrt 2 $

Hence the equation of required circle is

${(x - 1)^2} + {(y - 2)^2} = {(\sqrt 2 )^2}$

$ \Rightarrow \,{x^2} + {y^2} - 2x - 4y + 3 = 0.$

The equation of circle with centre $(1, 2)$ and tangent $x + y - 5 = 0$ is

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