The slope of one of the direct common tangent | vedclass

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(B) Given circles are $C_1: x^2+y^2-2x+4y+1=0$ and $C_2: x^2+y^2-4x-2y+4=0$. Centers are $C_1(1, -2)$ and $C_2(2, 1)$. Radii are $r_1 = \sqrt{1^2+(-2)

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(B) Given circles are $C_1: x^2+y^2-2x+4y+1=0$ and $C_2: x^2+y^2-4x-2y+4=0$. Centers are $C_1(1, -2)$ and $C_2(2, 1)$. Radii are $r_1 = \sqrt{1^2+(-2)^2-1} = 2$ and $r_2 = \sqrt{2^2+1^2-4} = 1$. Let the common tangent be $y-mx-c=0$. The distance from center to the tangent equals the radius: $\frac{|-2-m(1)-c|}{\sqrt{1+m^2}} = 2$ and $\frac{|1-m(2)-c|}{\sqrt{1+m^2}} = 1$. For direct common tangents,the centers lie on the same side,so $\frac{-2-m-c}{\sqrt{1+m^2}} = 2$ and $\frac{1-2m-c}{\sqrt{1+m^2}} = 1$. Subtracting the equations: $\frac{-3+m}{\sqrt{1+m^2}} = 1$. Squaring both sides: $m^2-6m+9 = 1+m^2$,which gives $6m = 8$,so $m = \frac{4}{3}$.

The slope of one of the direct common tangents drawn to the circles $x^2+y^2-2x+4y+1=0$ and $x^2+y^2-4x-2y+4=0$ is

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