If the foot of the normal from the point (4,3 | vedclass

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(C) The normal at any point on a circle always passes through its center. The normal passes through $(4,3)$ and $(2,1)$.The equation of this normal is

Vedclass · Accepted Answer

$x^2+y^2-2x-1=0$

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(C) The normal at any point on a circle always passes through its center. The normal passes through $(4,3)$ and $(2,1)$.The equation of this normal is $y-1 = \frac{3-1}{4-2}(x-2)$ $\Rightarrow y-1 = 1(x-2)$ $\Rightarrow y = x-1 \dots(1)$.The center of the circle lies on this normal. We are also given that $2x-y-2=0$ is a diameter,so the center also lies on this line $\dots(2)$.Solving $(1)$ and $(2)$ by substituting $y = x-1$ into $2x-(x-1)-2=0$,we get $x-1=0 \Rightarrow x=1$. Then $y=0$. Thus,the center is $(1,0)$.The radius $r$ is the distance between the center $(1,0)$ and the point $(2,1)$ on the circle: $r = \sqrt{(2-1)^2 + (1-0)^2} = \sqrt{1^2+1^2} = \sqrt{2}$.The equation of the circle with center $(1,0)$ and radius $\sqrt{2}$ is $(x-1)^2 + (y-0)^2 = (\sqrt{2})^2$ $\Rightarrow x^2-2x+1+y^2=2$ $\Rightarrow x^2+y^2-2x-1=0$.

If the foot of the normal from the point $(4,3)$ to a circle is $(2,1)$ and $2x-y-2=0$ is a diameter of the circle,then the equation of the circle is

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