The slope of a common tangent to the circles | vedclass

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(D) Let the equation of the tangent be $y = mx + c$. For the circle $x^2 + y^2 = 16$,the radius is $r = 4$ and the center is $(0, 0)$. The condition f

Vedclass · Accepted Answer

$\frac{8}{\sqrt{17}}$

Vedclass · Answer

(D) Let the equation of the tangent be $y = mx + c$. For the circle $x^2 + y^2 = 16$,the radius is $r = 4$ and the center is $(0, 0)$. The condition for the line $mx - y + c = 0$ to be a tangent is $\frac{|c|}{\sqrt{m^2 + 1}} = 4$,so $c^2 = 16(m^2 + 1)$. For the circle $(x - 9)^2 + y^2 = 16$,the center is $(9, 0)$ and the radius is $r = 4$. The condition for the line $mx - y + c = 0$ to be a tangent is $\frac{|9m + c|}{\sqrt{m^2 + 1}} = 4$,so $(9m + c)^2 = 16(m^2 + 1)$. Equating the two expressions for $16(m^2 + 1)$,we get $c^2 = (9m + c)^2$. This implies $c = -(9m + c)$ or $c = 9m + c$. Case $1$: $c = 9m + c \implies 9m = 0 \implies m = 0$. Case $2$: $c = -9m - c \implies 2c = -9m \implies c = -\frac{9m}{2}$. Substituting $c = -\frac{9m}{2}$ into $c^2 = 16(m^2 + 1)$: $\frac{81m^2}{4} = 16m^2 + 16 \implies 81m^2 = 64m^2 + 64 \implies 17m^2 = 64 \implies m^2 = \frac{64}{17}$. Thus,$m = \pm \frac{8}{\sqrt{17}}$. Comparing with the options,the correct slope is $\frac{8}{\sqrt{17}}$.

The slope of a common tangent to the circles $x^2+y^2=16$ and $(x-9)^2+y^2=16$ is

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