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

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(A) The condition for the line $y = mx + c$ to be a tangent to the circle $x^2 + y^2 = a^2$ is $c^2 = a^2(1 + m^2)$.Substituting $y = mx + c$ into the

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$\left( \frac{-a^2m}{c}, \frac{a^2}{c} \right)$

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(A) The condition for the line $y = mx + c$ to be a tangent to the circle $x^2 + y^2 = a^2$ is $c^2 = a^2(1 + m^2)$.Substituting $y = mx + c$ into the circle equation: $x^2 + (mx + c)^2 = a^2$.$x^2 + m^2x^2 + 2mxc + c^2 = a^2$.$(1 + m^2)x^2 + 2mxc + (c^2 - a^2) = 0$.Since the line is a tangent,the discriminant must be zero,leading to the x-coordinate of the point of contact: $x = -\frac{2mc}{2(1 + m^2)} = -\frac{mc}{1 + m^2}$.Using $c^2 = a^2(1 + m^2)$,we have $1 + m^2 = \frac{c^2}{a^2}$.Thus,$x = -\frac{mc}{c^2/a^2} = -\frac{a^2m}{c}$.Substituting $x$ back into $y = mx + c$: $y = m(-\frac{a^2m}{c}) + c = \frac{-a^2m^2 + c^2}{c} = \frac{a^2(1 + m^2) - a^2m^2}{c} = \frac{a^2}{c}$.Therefore,the point of contact is $\left( -\frac{a^2m}{c}, \frac{a^2}{c} \right)$.

If the line $y = mx + c$ is a tangent to the circle $x^2 + y^2 = a^2$,then the point of contact is

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