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

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Question

(B) The radius of the circle is the perpendicular distance from the centre $(1, 2)$ to the tangent line $x + y - 5 = 0$.$r = \frac{|1 + 2 - 5|}{\sqrt{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The radius of the circle is the perpendicular distance from the centre $(1, 2)$ to the tangent line $x + y - 5 = 0$.$r = \frac{|1 + 2 - 5|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$The equation of the circle with centre $(h, k) = (1, 2)$ and radius $r = \sqrt{2}$ is given by $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y - 2)^2 = (\sqrt{2})^2$$(x^2 - 2x + 1) + (y^2 - 4y + 4) = 2$$x^2 + y^2 - 2x - 4y + 5 = 2$$x^2 + y^2 - 2x - 4y + 3 = 0$Thus,the correct option is $B$.

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

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