Q60 / 90 · JEE Main 2010 · Mathematics

The circle $x^2+y^2=4 x+8 y+5$ intersects the line $3 x-4 y=m$ at two distinct points if

  • Correct answer
    A
    $-35 < m < 15$
  • B
    $15 < m < 65$
  • C
    $35 < m < 85$
  • D
    $-85 < m < -35$

Solution

Circle $x^2+y^2-4 x-8 y-5=0$ Centre $=(2,4)$, Radius $=\sqrt{4+16+5}=5$ If circle is intersecting line $3 x-4 y=m$ at two distinct points. $\Rightarrow$ length of perpendicular from centre $ < $ radius $\Rightarrow \frac{|6-16-\mathrm{m}|}{5} < 5$ $\Rightarrow|10+\mathrm{m}| < 25$ $\Rightarrow-25 < m+10 < 25$ $\Rightarrow-35 < \mathrm{m} < 15$.

Hence, the correct answer is option: 'A'.

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