If {c^2} {a^2}(1 + {m^2}),then the line y = | vedclass

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(C) The condition for the line $y = mx + c$ to be a tangent to the circle ${x^2} + {y^2} = {a^2}$ is ${c^2} = {a^2}(1 + {m^2})$.If ${c^2} < {a^2}(1 +

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At no point

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(C) The condition for the line $y = mx + c$ to be a tangent to the circle ${x^2} + {y^2} = {a^2}$ is ${c^2} = {a^2}(1 + {m^2})$.If ${c^2} If ${c^2} > {a^2}(1 + {m^2})$,the distance from the center $(0, 0)$ to the line $mx - y + c = 0$ is given by $d = \frac{|c|}{\sqrt{m^2 + 1}}$.Since ${c^2} > {a^2}(1 + {m^2})$,we have $\frac{|c|}{\sqrt{m^2 + 1}} > |a|$.This means the perpendicular distance from the center to the line is greater than the radius of the circle.Therefore,the line does not intersect the circle at any point.

If ${c^2} > {a^2}(1 + {m^2})$,then the line $y = mx + c$ will intersect the circle ${x^2} + {y^2} = {a^2}$ at:

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