The circumference of a circle passing through | vedclass

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(B) The intersection point of two normals to a circle is the center of the circle.Given equations of normals are:$2x - 3y + 4 = 0$ ... $(i)$$x + y - 3

Vedclass · Accepted Answer

$10$

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(B) The intersection point of two normals to a circle is the center of the circle.Given equations of normals are:$2x - 3y + 4 = 0$ ... $(i)$$x + y - 3 = 0$ ... $(ii)$Multiplying $(ii)$ by $2$,we get $2x + 2y - 6 = 0$ ... $(iii)$Subtracting $(i)$ from $(iii)$:$(2x + 2y - 6) - (2x - 3y + 4) = 0$$5y - 10 = 0 \implies y = 2$Substituting $y = 2$ in $(ii)$:$x + 2 - 3 = 0 \implies x = 1$So,the center of the circle is $(1, 2)$.The circle passes through the point $(4, 6)$.The radius $r$ is the distance between $(1, 2)$ and $(4, 6)$:$r = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.The circumference of the circle is $2 \pi r = 2 \pi (5) = 10 \pi$.

The circumference of a circle passing through the point $(4,6)$ with two normals represented by $2x - 3y + 4 = 0$ and $x + y - 3 = 0$ is (in $\pi$)

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