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

Q: The straight line y = mx + c cuts the circle x2 + y2 = a2 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 x2 + y2  = a2 in real points, then the equation , x2 +(mx + c)2 = a2

i.e. x2 (1+ m2) + 2mcx + c2 – a2 = 0 has real roots.

Hence , 4m2 c2 –4 (1 + m2) (c2 – a2)  ≥ 0

or  –c2 + a2 (1+m2)  ≥ 0

or , a2(1 + m2) ≥ c2

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

Hence (C) is the correct answer.