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

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Question

(A) The given equation of the circle is $x^2+y^2-16x-12y+64=0$. The center is $(8, 6)$ and the radius is $r = \sqrt{8^2+6^2-64} = \sqrt{64+36-64} = 6$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given equation of the circle is $x^2+y^2-16x-12y+64=0$. The center is $(8, 6)$ and the radius is $r = \sqrt{8^2+6^2-64} = \sqrt{64+36-64} = 6$.$(i)$ The equation of the polar of $(x_1, y_1)$ w.r.t. $x^2+y^2+2gx+2fy+c=0$ is $xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0$.For $(-5, 1)$,it is $x(-5)+y(1)-8(x-5)-6(y+1)+64=0$ $\Rightarrow -5x+y-8x+40-6y-6+64=0$ $\Rightarrow -13x-5y+98=0$ $\Rightarrow 13x+5y=98$. 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 $(8, 0)$,it is $x(8)+y(0)-8(x+8)-6(y+0)+64=0$ $\Rightarrow 8x-8x-64-6y+64=0$ $\Rightarrow -6y=0$ $\Rightarrow y=0$. This matches $(A)$.$(iii)$ The normal at any point on a circle passes through the center $(8, 6)$. The normal at $(2, 6)$ passes through $(2, 6)$ and $(8, 6)$. Since the $y$-coordinates are the same,the equation is $y=6$. This matches $(B)$.$(iv)$ The diameter passes through the center $(8, 6)$ and the point $(8, 12)$. Since the $x$-coordinates are the same,the equation 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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