Let C be the circle of minimum area touching | vedclass

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Question

(A) Let the center of the circle be $(0, k)$ and its radius be $r$. Since the circle is symmetric about the $y$-axis and touches the parabola $y=6-x^2

Vedclass · Accepted Answer

$(2,4)$

Vedclass · Answer

(A) Let the center of the circle be $(0, k)$ and its radius be $r$. Since the circle is symmetric about the $y$-axis and touches the parabola $y=6-x^2$ at its vertex $(0, 6)$,the center must be at $(0, 6-r)$.The equation of the circle is $x^2 + (y-(6-r))^2 = r^2$.The circle also touches the lines $y = \sqrt{3}x$ and $y = -\sqrt{3}x$,which can be written as $\sqrt{3}x - y = 0$ and $\sqrt{3}x + y = 0$.The perpendicular distance from the center $(0, 6-r)$ to the line $\sqrt{3}x - y = 0$ must be equal to the radius $r$:$\frac{|\sqrt{3}(0) - (6-r)|}{\sqrt{(\sqrt{3})^2 + (-1)^2}} = r$$\frac{|r-6|}{\sqrt{3+1}} = r$$\frac{|r-6|}{2} = r$$|r-6| = 2r$Since the circle is below the vertex,$r The equation of the circle is $x^2 + (y-(6-2))^2 = 2^2$,which simplifies to $x^2 + (y-4)^2 = 4$.Checking the given points:For $(2, 4)$,$2^2 + (4-4)^2 = 4 + 0 = 4$. Thus,$(2, 4)$ lies on the circle.

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 $(1, 3)$	$IV$. $x=9$

Let $C$ be the circle of minimum area touching the parabola $y=6-x^2$ and the lines $y=\sqrt{3}|x|$. Then,which one of the following points lies on the circle $C$?

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