The combined equation of the direct common ta | vedclass

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(D) For circle $C_1: x^2+y^2-2x-2y-2=0$,center $O_1(1, 1)$ and radius $r_1 = \sqrt{1^2+1^2-(-2)} = 2$. For circle $C_2: x^2+y^2+4x+6y+12=0$,center $O_

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$15x^2-24xy+8y^2-18x-8y-73=0$

Vedclass · Answer

(D) For circle $C_1: x^2+y^2-2x-2y-2=0$,center $O_1(1, 1)$ and radius $r_1 = \sqrt{1^2+1^2-(-2)} = 2$. For circle $C_2: x^2+y^2+4x+6y+12=0$,center $O_2(-2, -3)$ and radius $r_2 = \sqrt{(-2)^2+(-3)^2-12} = 1$. The direct common tangents intersect at the external center of similitude $S$,which divides $O_1O_2$ externally in the ratio $r_1:r_2 = 2:1$. $S = \left(\frac{2(-2)-1(1)}{2-1}, \frac{2(-3)-1(1)}{2-1}\right) = (-5, -7)$. Any line through $S(-5, -7)$ is $y+7 = m(x+5) \implies mx-y+5m-7=0$. The distance from $O_1(1, 1)$ to this line is equal to $r_1=2$: $\frac{|m(1)-1+5m-7|}{\sqrt{m^2+1}} = 2 \implies |6m-8| = 2\sqrt{m^2+1} \implies |3m-4| = \sqrt{m^2+1}$. Squaring both sides: $9m^2-24m+16 = m^2+1 \implies 8m^2-24m+15=0$. The combined equation of the tangents is $(y+7-m(x+5))(y+7-m'(x+5)) = 0$,where $m, m'$ are roots of $8m^2-24m+15=0$. Let $Y = y+7$ and $X = x+5$. Then $m^2X^2 - mX(Y) + Y^2 = 0$ is not the form; rather,the equation is $(Y-mX)(Y-m'X) = Y^2 - (m+m')XY + mm'X^2 = 0$. Here $m+m' = 3$ and $mm' = 15/8$. $Y^2 - 3XY + \frac{15}{8}X^2 = 0 \implies 15X^2 - 24XY + 8Y^2 = 0$. Substituting $X=x+5, Y=y+7$: $15(x+5)^2 - 24(x+5)(y+7) + 8(y+7)^2 = 0$. $15(x^2+10x+25) - 24(xy+7x+5y+35) + 8(y^2+14y+49) = 0$. $15x^2+150x+375 - 24xy-168x-120y-840 + 8y^2+112y+392 = 0$. $15x^2-24xy+8y^2-18x-8y-73=0$.

The combined equation of the direct common tangents of the circles $x^2+y^2-2x-2y-2=0$ and $x^2+y^2+4x+6y+12=0$ is:

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