If x-2y=0 is a tangent drawn at a point P on | vedclass

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(B) The equation of the circle is $x^2+y^2-6x+2y+c=0$. The center of the circle is $C = (3, -1)$.Since $x-2y=0$ is a tangent at point $P$,the radius $

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$2\sqrt{5}$

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(B) The equation of the circle is $x^2+y^2-6x+2y+c=0$. The center of the circle is $C = (3, -1)$.Since $x-2y=0$ is a tangent at point $P$,the radius $CP$ is perpendicular to the tangent line.The slope of the tangent $x-2y=0$ is $m_1 = 1/2$. Thus,the slope of the normal $CP$ is $m_2 = -2$.The equation of the normal line passing through $C(3, -1)$ with slope $-2$ is $y - (-1) = -2(x - 3)$,which simplifies to $y+1 = -2x+6$,or $2x+y-5=0$.Point $P$ is the intersection of the tangent $x-2y=0$ and the normal $2x+y-5=0$.Solving these: $x=2y$,so $2(2y)+y-5=0 \implies 5y=5 \implies y=1$. Then $x=2(1)=2$. So $P = (2, 1)$.The distance between $P(2, 1)$ and the point $(6, 3)$ is given by the distance formula:$d = \sqrt{(6-2)^2 + (3-1)^2} = \sqrt{4^2 + 2^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.

If $x-2y=0$ is a tangent drawn at a point $P$ on the circle $x^2+y^2-6x+2y+c=0$,then the distance of the point $(6,3)$ from $P$ is

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