The equation of a circle which has a tangent | vedclass

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(C) The normals to a circle intersect at its center. Given $(x - 1)(y - 2) = 0$,the center of the circle is $(1, 2)$.Since the line $3x + 4y = 6$ is a

Vedclass · Accepted Answer

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

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(C) The normals to a circle intersect at its center. Given $(x - 1)(y - 2) = 0$,the center of the circle is $(1, 2)$.Since the line $3x + 4y = 6$ is a tangent to the circle,the radius $r$ is the perpendicular distance from the center $(1, 2)$ to the line $3x + 4y - 6 = 0$.$r = \frac{|3(1) + 4(2) - 6|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 6|}{\sqrt{25}} = \frac{5}{5} = 1$.The equation of the circle with center $(h, k) = (1, 2)$ and radius $r = 1$ is $(x - h)^2 + (y - k)^2 = r^2$.$(x - 1)^2 + (y - 2)^2 = 1^2$.$x^2 - 2x + 1 + y^2 - 4y + 4 = 1$.$x^2 + y^2 - 2x - 4y + 4 = 0$.

The equation of a circle which has a tangent $3x + 4y = 6$ and two normals given by $(x - 1)(y - 2) = 0$ is

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