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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

If the circles \({x^2}\, + {y^2}\, - 16x\, - 20y\, + \,164\,\, = \,\,{r^2}\) and \({(x - 4)^2} + {(y - 7)^2} = 36\) intersect at two distinct points, then

  1. A \(0 < r < 1\)
  2. B \(1 < r < 11\)
  3. C \(r>11\)
  4. D \(r=11\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(1 < r < 11\)

Step-by-step Solution

Detailed explanation

\({x^2} + {y^2} - 16x - 20y + 164 = {r^2}\) \(A\left( {8,10} \right),{R_1} = r\) \({\left( {x - 4} \right)^2} + {\left( {y - 7} \right)^2} = 36\) \(B\left( {4,7,} \right),{R_2} = 6\) \(\left| {{R_1} - {R_2}} \right| < AB < {R_1} + {R_2}\) \( \Rightarrow 1 < r < 11\)
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