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

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Question

(C) The centre of the circle is $(h, k) = (2, 1)$.The radius $r$ is the perpendicular distance from the centre $(2, 1)$ to the line $3x + 4y - 5 = 0$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The centre of the circle is $(h, k) = (2, 1)$.The radius $r$ is the perpendicular distance from the centre $(2, 1)$ to the line $3x + 4y - 5 = 0$.Using the formula $r = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$,we get:$r = \frac{|3(2) + 4(1) - 5|}{\sqrt{3^2 + 4^2}} = \frac{|6 + 4 - 5|}{\sqrt{9 + 16}} = \frac{5}{5} = 1$.The equation of the circle is $(x - h)^2 + (y - k)^2 = r^2$.Substituting the values: $(x - 2)^2 + (y - 1)^2 = 1^2$.Expanding this: $(x^2 - 4x + 4) + (y^2 - 2y + 1) = 1$.$x^2 + y^2 - 4x - 2y + 5 = 1$.$x^2 + y^2 - 4x - 2y + 4 = 0$.

The equation of the circle with centre $(2, 1)$ and touching the line $3x + 4y = 5$ is

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