Given the circle C with the equation x^2+y^2- | vedclass

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(A) The given equation of the circle is $x^2+y^2-2x+10y-38=0$.$(A)$ The equation of the polar of $(x_1, y_1)$ with respect to $x^2+y^2+2gx+2fy+c=0$ is

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$A-III, B-IV, C-I, D-II$

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(A) The given equation of the circle is $x^2+y^2-2x+10y-38=0$.$(A)$ The equation of the polar of $(x_1, y_1)$ with respect to $x^2+y^2+2gx+2fy+c=0$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.For $(4, 3)$,we have $x(4)+y(3)-1(x+4)+5(y+3)-38=0$.$4x+3y-x-4+5y+15-38=0$ $\Rightarrow 3x+8y-27=0$ $\Rightarrow 3x+8y=27$. Thus,$A-III$.$(B)$ The equation of the tangent at $(x_1, y_1)$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.For $(9, -5)$,we have $x(9)+y(-5)-1(x+9)+5(y-5)-38=0$.$9x-5y-x-9+5y-25-38=0$ $\Rightarrow 8x-72=0$ $\Rightarrow x=9$. Thus,$B-IV$.$(C)$ The centre of the circle is $(-g, -f) = (1, -5)$. The normal at any point on the circle passes through the centre.The slope of the normal at $(-7, -5)$ is the slope of the line joining $(-7, -5)$ and $(1, -5)$.Slope $m = \frac{-5-(-5)}{1-(-7)} = \frac{0}{8} = 0$.The equation of the normal is $y-(-5) = 0(x-(-7)) \Rightarrow y+5=0$. Thus,$C-I$.$(D)$ The diameter passes through the centre $(1, -5)$ and the point $(1, 3)$.Since both points have the same $x$-coordinate,the equation of the line is $x=1$. Thus,$D-II$.Therefore,the correct matching is $A-III, B-IV, C-I, D-II$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

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