If y+c=0 is a tangent to the circle x^2+y^2-6 | vedclass

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(B) The given equation of the circle is $x^2+y^2-6x-2y+1=0$. Since the point $(a, 4)$ lies on the circle,we substitute $y=4$ into the circle equation:

Vedclass · Accepted Answer

$ac=-12$

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(B) The given equation of the circle is $x^2+y^2-6x-2y+1=0$. Since the point $(a, 4)$ lies on the circle,we substitute $y=4$ into the circle equation: $x^2+4^2-6x-2(4)+1=0$ $x^2-6x+16-8+1=0$ $x^2-6x+9=0$ $(x-3)^2=0 \Rightarrow x=3$. Thus,the point of tangency is $(3, 4)$,so $a=3$. The equation of the tangent at $(x_1, y_1)$ to the circle $x^2+y^2+2gx+2fy+c'=0$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c'=0$. Here,$g=-3, f=-1, c'=1$ and $(x_1, y_1)=(3, 4)$. Substituting these values: $x(3)+y(4)-3(x+3)-1(y+4)+1=0$ $3x+4y-3x-9-y-4+1=0$ $3y-12=0 \Rightarrow y-4=0$. Comparing this with $y+c=0$,we get $c=-4$. Therefore,$ac = 3 	imes (-4) = -12$.

If $y+c=0$ is a tangent to the circle $x^2+y^2-6x-2y+1=0$ at $(a, 4)$,then

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