Find the equation of the circle having normal | vedclass

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Question

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

Vedclass · Accepted Answer

$(x-1)^2+(y-2)^2=1$

Vedclass · Answer

(A) The normals to a circle always intersect at its center. Given the normals $(x-1)(y-2)=0$,the center of the circle is $(1, 2)$.Since $3x+4y=6$ is a tangent to the circle,the radius $r$ is the perpendicular distance from the center $(1, 2)$ to the line $3x+4y-6=0$.$r = \frac{|3(1) + 4(2) - 6|}{\sqrt{3^2 + 4^2}} = \frac{|3 + 8 - 6|}{\sqrt{9 + 16}} = \frac{5}{5} = 1$.The equation of the circle with center $(h, k) = (1, 2)$ and radius $r = 1$ is $(x-h)^2 + (y-k)^2 = r^2$.$(x-1)^2 + (y-2)^2 = 1^2$,which simplifies to $(x-1)^2 + (y-2)^2 = 1$.Thus,option $A$ is correct.

Find the equation of the circle having normals $(x-1)(y-2)=0$ and a tangent $3x+4y=6$.

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