The set of all real values of lambda for whi | vedclass

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Question

The equation of the circles are

${x^2} + {y^2} - 10x - 10y + \lambda  = 0\,\,\,\,\,\,\,......\left( 1 \right)$

and ${x^2} + {y^2} - 4x

Vedclass · Accepted Answer

$(18, 42)$

Vedclass · Answer

The equation of the circles are

${x^2} + {y^2} - 10x - 10y + \lambda  = 0\,\,\,\,\,\,\,......\left( 1 \right)$

and ${x^2} + {y^2} - 4x - 4y + 6 = 0\,\,\,\,\,\,\,......\left( 2 \right)$

${C_1} = \,$ center of $\left( 1 \right) = \left( {5,5} \right)$

${C_2} = \,$ center of $\,\left( 2 \right) = \left( {2,2} \right)$

$ = {C_1}{C_2} = \sqrt {9 + 9}  = \sqrt {18} $

${r_1} = \sqrt {50 - \lambda } ,{r_2} = \sqrt 2 $

For exactly two common tangents we have

${r_1} - {r_2} < {C_1}{C_2} < {r_1} + {r_2}$

$ \Rightarrow \sqrt {50 - \lambda }  - \sqrt 2  < 3\sqrt 2  < \sqrt {50 - \lambda }  + \sqrt 2 $

$ \Rightarrow \sqrt {50 - \lambda }  - \sqrt 2  < 3\sqrt 2 $ or $\,3\sqrt 2  < \sqrt {50 - \lambda }  + \sqrt 2 $

$ \Rightarrow \sqrt {50 - \lambda }  < 4\sqrt 2 $ or $2\sqrt 2  < \sqrt {50 - \lambda } $

$50 - \lambda  < 32$ or $8 > 50 - \lambda $

$ \Rightarrow \lambda  < 18$ or $\lambda  < 42$

Required interval is $(18,42)$

The set of all real values of $\lambda $ for which exactly two common tangents can be drawn to the circles $x^2 + y^2 - 4x - 4y+ 6\, = 0$ and $x^2 + y^2 - 10x - 10y + \lambda \, = 0$ is the interval:

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