If the line y=mx+c is a common tangent to the | vedclass

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(B) The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$.For t

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$4c^{2}=369$

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(B) The condition for the line $y=mx+c$ to be a tangent to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $c^{2}=a^{2}m^{2}-b^{2}$.For the given hyperbola $\frac{x^{2}}{100}-\frac{y^{2}}{64}=1$,we have $a^{2}=100$ and $b^{2}=64$,so $c^{2}=100m^{2}-64$.The condition for the line $y=mx+c$ to be a tangent to the circle $x^{2}+y^{2}=r^{2}$ is $c^{2}=r^{2}(1+m^{2})$.For the given circle $x^{2}+y^{2}=36$,we have $r^{2}=36$,so $c^{2}=36(1+m^{2})$.Equating the two expressions for $c^{2}$:$100m^{2}-64=36(1+m^{2})$$100m^{2}-64=36+36m^{2}$$64m^{2}=100$$m^{2}=\frac{100}{64}=\frac{25}{16}$.Now,substitute $m^{2}$ back into the circle's condition:$c^{2}=36(1+\frac{25}{16})$$c^{2}=36(\frac{16+25}{16})$$c^{2}=36(\frac{41}{16})$$c^{2}=\frac{9 	imes 41}{4}$$4c^{2}=369$.

If the line $y=mx+c$ is a common tangent to the hyperbola $\frac{x^{2}}{100}-\frac{y^{2}}{64}=1$ and the circle $x^{2}+y^{2}=36,$ then which one of the following is true?

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