The circles x^2 + y^2 + 4x + 6y + 3 = 0 and 2 | vedclass

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(B) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.First,rewrite the equations in standard form:Circle $1$: $x^2 + y^2 + 4x + 6y + 3 = 0$. Here,$2g_1 = 4 \Rightarrow g_1 = 2$,$2f_1 = 6 \Rightarrow f_1 = 3$,and $c_1 = 3$.Circle $2$: $2(x^2 + y^2) + 6x + 4y + C = 0 \Rightarrow x^2 + y^2 + 3x + 2y + \frac{C}{2} = 0$. Here,$2g_2 = 3 \Rightarrow g_2 = \frac{3}{2}$,$2f_2 = 2 \Rightarrow f_2 = 1$,and $c_2 = \frac{C}{2}$.Applying the condition: $2(2)(\frac{3}{2}) + 2(3)(1) = 3 + \frac{C}{2}$.$6 + 6 = 3 + \frac{C}{2}$ $\Rightarrow 12 = 3 + \frac{C}{2}$ $\Rightarrow 9 = \frac{C}{2}$ $\Rightarrow C = 18$.

List-$I$	List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$	$I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$	$II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$	$III.$ $60^{\circ}$
	$IV.$ $30^{\circ}$

The circles $x^2 + y^2 + 4x + 6y + 3 = 0$ and $2(x^2 + y^2) + 6x + 4y + C = 0$ will cut orthogonally,if $C$ equals

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