A) 4
B) 5
C) 6
D) 7
E) 8
Correct Answer: E
Solution :
\[\because \] (4, 7) and\[(-2,-1)\]are end points of circle. \[\Rightarrow \]Centre of circle \[=\left( \frac{4-2}{2},\frac{7-1}{2} \right)=(1,3)\] \[\therefore \]Slope of\[AC=\frac{-1-0}{-2-a}=-\frac{1}{2+a}\] and slope of\[AD=\frac{7-0}{4-a}=\frac{7}{4-a}\] Since,\[\Delta CAD\]is right angle as CD is diameter. \[\Rightarrow \](slope of AC (slope of AD)\[=-1\] \[\Rightarrow \] \[\left( \frac{1}{2+a} \right)\left( \frac{7}{4-a} \right)=-1\] \[\Rightarrow \] \[\frac{7}{-{{a}^{2}}+2a+8}=-1\] \[\Rightarrow \] \[{{a}^{2}}-2a-15=0\] \[\Rightarrow \] \[a=-3,5\] Similarly, \[b=-3,5\] \[\therefore \]Point\[A=(-3,0)\]and point\[B=(5,0)\] \[\therefore \] \[AB=\sqrt{{{(-3-5)}^{2}}+{{(0-0)}^{2}}}\] \[=\sqrt{{{(-8)}^{2}}}=8\]You need to login to perform this action.
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