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(C) Let the equation of the circle be ${x^2} + {y^2} + 2gx + 2fy = 0$ since it passes through the origin $(0, 0)$.Since the centre $(-g, -f)$ lies on

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${x^2} + {y^2} - 4x - 4y = 0$

Vedclass · Answer

(C) Let the equation of the circle be ${x^2} + {y^2} + 2gx + 2fy = 0$ since it passes through the origin $(0, 0)$.Since the centre $(-g, -f)$ lies on the line $x + y = 4$,we have $-g - f = 4$,or $g + f = -4$ ... $(i)$.The circle cuts ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally. The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = g, f_1 = f, c_1 = 0$ and $g_2 = -2, f_2 = 1, c_2 = 4$.So,$2(g)(-2) + 2(f)(1) = 0 + 4$,which simplifies to $-4g + 2f = 4$,or $-2g + f = 2$ ... $(ii)$.Subtracting $(ii)$ from $(i)$: $(g + f) - (-2g + f) = -4 - 2$,which gives $3g = -6$,so $g = -2$.Substituting $g = -2$ into $(i)$: $-2 + f = -4$,so $f = -2$.The equation of the circle is ${x^2} + {y^2} + 2(-2)x + 2(-2)y = 0$,which is ${x^2} + {y^2} - 4x - 4y = 0$.

List-$I$	List-$II$
$A$. $x^2+y^2+2x+8y-23=0$,$x^2+y^2-4x-10y+19=0$	$I$. $0$
$B$. $x^2+y^2=1$,$x^2+y^2-2x-6y+6=0$	$II$. $1$
$C$. $x^2+y^2-8x+2y=0$,$x^2+y^2-2x-16y+25=0$	$III$. $2$
$D$. $x^2+y^2=4$,$x^2+y^2-2x=0$	$IV$. $3$
	$V$. $4$

The equation of the circle which passes through the origin,has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally,is

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