Consider the curves C_1: y^2=4x and C_2: x^2+ | vedclass

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(A) Given: $C_1: y^2=4x$,where $a=1$. $C_2: x^2+y^2-6x+1=0$,which has center $(3,0)$ and radius $r = \sqrt{3^2+0^2-1} = \sqrt{8} = 2\sqrt{2}$. Equatio

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Both Assertion and Reason are true and Reason is the correct explanation of Assertion.

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(A) Given: $C_1: y^2=4x$,where $a=1$. $C_2: x^2+y^2-6x+1=0$,which has center $(3,0)$ and radius $r = \sqrt{3^2+0^2-1} = \sqrt{8} = 2\sqrt{2}$. Equation of tangent to $C_1$ in slope form is $y = mx + \frac{1}{m}$,or $m^2x - my + 1 = 0$. Since this is also a tangent to $C_2$,the perpendicular distance from the center $(3,0)$ to this line must equal the radius $2\sqrt{2}$. $\frac{|m^2(3) - m(0) + 1|}{\sqrt{(m^2)^2 + (-m)^2}} = 2\sqrt{2}$ $\frac{|3m^2+1|}{\sqrt{m^4+m^2}} = 2\sqrt{2}$ Squaring both sides: $(3m^2+1)^2 = 8(m^4+m^2)$ $9m^4 + 6m^2 + 1 = 8m^4 + 8m^2$ $m^4 - 2m^2 + 1 = 0$ $(m^2-1)^2 = 0 \Rightarrow m = \pm 1$. For $m=1$,the tangent is $y = x + 1 \Rightarrow x - y + 1 = 0$. For $m=-1$,the tangent is $y = -x - 1 \Rightarrow x + y + 1 = 0$. Since the slopes are $m_1=1$ and $m_2=-1$,their product $m_1 \cdot m_2 = -1$,which means the tangents are orthogonal. Thus,both Assertion and Reason are true,and the Reason correctly explains the Assertion.

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

Consider the curves $C_1: y^2=4x$ and $C_2: x^2+y^2-6x+1=0$. Assertion $(A)$: The common tangents to the curves $C_1$ and $C_2$ are orthogonal. Reason $(R)$: $x-y+1=0$ and $x+y+1=0$ are the common tangents to the curves $C_1$ and $C_2$.

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