The distance of the origin from the external | vedclass

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(A) Given circles are $S_1: x^2+y^2-8x-10y-8=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,center $C_1 = (4, 5)$ and radius $r_1 = \sqrt{4^2+5^2-(-8)} =

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$\frac{3 \sqrt{26}}{5}$

Vedclass · Answer

(A) Given circles are $S_1: x^2+y^2-8x-10y-8=0$ and $S_2: x^2+y^2+2x-2y-2=0$. For $S_1$,center $C_1 = (4, 5)$ and radius $r_1 = \sqrt{4^2+5^2-(-8)} = \sqrt{16+25+8} = \sqrt{49} = 7$. For $S_2$,center $C_2 = (-1, 1)$ and radius $r_2 = \sqrt{(-1)^2+1^2-(-2)} = \sqrt{1+1+2} = \sqrt{4} = 2$. The external centre of similitude $Q$ divides the line segment joining $C_1$ and $C_2$ externally in the ratio $r_1 : r_2$. $Q = \left( \frac{r_1 x_2 - r_2 x_1}{r_1 - r_2}, \frac{r_1 y_2 - r_2 y_1}{r_1 - r_2} ight) = \left( \frac{7(-1) - 2(4)}{7-2}, \frac{7(1) - 2(5)}{7-2} ight) = \left( \frac{-7-8}{5}, \frac{7-10}{5} ight) = \left( -3, -\frac{3}{5} ight)$. The distance of $Q$ from the origin $(0,0)$ is $D = \sqrt{(-3)^2 + (-3/5)^2} = \sqrt{9 + 9/25} = \sqrt{\frac{225+9}{25}} = \sqrt{\frac{234}{25}} = \frac{\sqrt{9 	imes 26}}{5} = \frac{3\sqrt{26}}{5}$.

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 distance of the origin from the external centre of similitude for the circles $x^2+y^2-8x-10y-8=0$ and $x^2+y^2+2x-2y-2=0$ is

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