A) 7 and 12
B) Only 7
C) Only 12
D) None of these
Correct Answer: A
Solution :
Let circle through (1, 2) and (3, 4) as ends of diameter be \[(x-1)(x-3)+(y-2)(y-4)=0\] and line through these points is\[(y-2)=\left( \frac{4-2}{3-1} \right)\text{ }(x-1)\]. Family of circles through the points and the line are \[(x-1)(x-3)+(y-2)(y-4)+\lambda (y-x-1)=0\]. \[\Rightarrow {{x}^{2}}+{{y}^{2}}+x(-4-\lambda )+y(-6+\lambda )+(3+8-\lambda )=0\] This circle touches \[3x+y-3=0\]. Now, radius = perpendicular distance between the centre and the line. \[\Rightarrow \sqrt{{{\left( \frac{4+\lambda }{2} \right)}^{2}}+{{\left( \frac{6-\lambda }{2} \right)}^{2}}-(3+8-\lambda )}\] \[=\left| \frac{3\text{ }\left( \frac{4+\lambda }{2} \right)+\left( \frac{6-\lambda }{2} \right)-3}{\sqrt{5}} \right|\] \[\Rightarrow \lambda =1,\ -4\Rightarrow c=7\] and 12.
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