For the circle C with the equation x^2+y^2-16 | vedclass

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(A) The given equation of the circle is $C: x^2 + y^2 - 16x - 12y + 64 = 0$.$(i)$ The equation of the polar of a point $(x_1, y_1)$ with respect to th

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$(D), (A), (B), (E)$

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(A) The given equation of the circle is $C: x^2 + y^2 - 16x - 12y + 64 = 0$.$(i)$ The equation of the polar of a point $(x_1, y_1)$ with respect to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is $xx_1 + yy_1 + g(x + x_1) + f(y + y_1) + c = 0$.For point $(-5, 1)$,$g = -8, f = -6, c = 64$:$x(-5) + y(1) - 8(x - 5) - 6(y + 1) + 64 = 0$$-5x + y - 8x + 40 - 6y - 6 + 64 = 0$$-13x - 5y + 98 = 0 \Rightarrow 13x + 5y = 98$. This matches $(D)$.$(ii)$ The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 - 8(x + x_1) - 6(y + y_1) + 64 = 0$.For $(8, 0)$:$8x + 0y - 8(x + 8) - 6(y + 0) + 64 = 0$$8x - 8x - 64 - 6y + 64 = 0$$-6y = 0 \Rightarrow y = 0$. This matches $(A)$.$(iii)$ The center of the circle is $(-g, -f) = (8, 6)$. The normal at any point on the circle passes through the center.The normal at $(2, 6)$ passes through $(2, 6)$ and $(8, 6)$.Since the $y$-coordinates are the same,the equation of the line is $y = 6$. This matches $(B)$.$(iv)$ The diameter passes through the center $(8, 6)$ and the given point $(8, 12)$.Since both points have the same $x$-coordinate,the equation of the line is $x = 8$. This matches $(E)$.Thus,the correct match is $(i)-(D), (ii)-(A), (iii)-(B), (iv)-(E)$.

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

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