If the two circles (x-1)^{2}+(y-3)^{2}=r^{2}\; and\; x^{2}+y^{2}-8x+2y+8=0    intersect in two distinct points, then

  • Option 1)

    r< 2\;

  • Option 2)

    \; r=2\;

  • Option 3)

    \; r> 2\;

  • Option 4)

    \; 2< r< 8

 

Answers (1)

As we learnt in

Common tangents of two circle -

When they intersect, there are two common tangents, both of them being direct.

- wherein

 

 (x-1)^{2}+(y-3)^{2}=r^{2}

x^{2}+y^{2}-8x+2y+8=0

C_{1}C_{2}=\sqrt{3^{2}+4^{2}}=5

r_{1}+r_{2}>5

r+3>5 \Rightarrow r>2

Also r - 3 < 5

r < 8

Thus 2 < r < 8

 


Option 1)

r< 2\;

Option 2)

\; r=2\;

Option 3)

\; r> 2\;

Option 4)

\; 2< r< 8

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