Each of the two orthogonal circles C_1 and C_ | vedclass

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(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+k=0$. Since the circle passes through $(2,0)$ and $(-2,0)$,we have $4+4g+k=0$ and $4-4g+k=0$. S

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$c^2=4(1+2m^2)$

Vedclass · Answer

(A) Let the equation of the circle be $x^2+y^2+2gx+2fy+k=0$. Since the circle passes through $(2,0)$ and $(-2,0)$,we have $4+4g+k=0$ and $4-4g+k=0$. Solving these gives $g=0$ and $k=-4$. Thus,the circles are of the form $x^2+y^2+2fy-4=0$. Let the two circles be $C_1: x^2+y^2+2f_1y-4=0$ and $C_2: x^2+y^2+2f_2y-4=0$. Since they are orthogonal,$2g_1g_2 + 2f_1f_2 = k_1+k_2$. Here $g_1=g_2=0$,so $2f_1f_2 = -4-4 = -8$,which implies $f_1f_2 = -4$. The line $y=mx+c$ is tangent to $x^2+y^2+2fy-4=0$,so the distance from the center $(0,-f)$ to the line $mx-y+c=0$ equals the radius $\sqrt{f^2+4}$. Thus,$|-f-c|/\sqrt{m^2+1} = \sqrt{f^2+4}$,which simplifies to $(f+c)^2 = (m^2+1)(f^2+4)$. Expanding gives $f^2+2cf+c^2 = m^2f^2+4m^2+f^2+4$,or $m^2f^2-2cf+4m^2+4-c^2=0$. Since $f_1$ and $f_2$ are roots of this quadratic in $f$,their product $f_1f_2 = (4m^2+4-c^2)/m^2$. Equating this to $-4$,we get $(4m^2+4-c^2)/m^2 = -4$,so $4m^2+4-c^2 = -4m^2$,which simplifies to $c^2 = 8m^2+4 = 4(1+2m^2)$.

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 $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

Each of the two orthogonal circles $C_1$ and $C_2$ passes through both the points $(2,0)$ and $(-2,0)$. If $y=mx+c$ is a common tangent to these circles,then

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