Q: The straight line y = mx + c cuts the circle x^{2} + y^{2} = a^{2} at real points if

(A) $\displaystyle \sqrt{a^2(1 + m^2) }\le |c|$

(B) $\displaystyle \sqrt{a^2(1 – m^2) }\le |c|$

(C) $\displaystyle \sqrt{a^2(1 + m^2) }\ge |c|$

(D) $\displaystyle \sqrt{a^2(1 – m^2) }\ge |c|$

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Ans: (C)

Sol: If the straight line y = mx + c ; cuts the circle x

^{2 }+ y^{2 }= a^{2}in real points, then the equation , x^{2}+(mx + c)^{2}= a^{2}i.e. x^{2} (1+ m^{2}) + 2mcx + c^{2} – a^{2} = 0 has real roots.

Hence , 4m^{2} c^{2} –4 (1 + m^{2}) (c^{2 }– a^{2}) ≥ 0

or –c^{2 }+ a^{2} (1+m^{2}) ≥ 0

or , a^{2}(1 + m^{2}) ≥ c^{2}

$\displaystyle \sqrt{a^2(1 + m^2) }\ge |c|$

Hence (C) is the correct answer.