question_answer

\[\Delta ABC\] is such that a circle touches AB at B and passes through centroid of \[\Delta ABC\] and C, if \[AB=6,\text{ }BC=4,\]then AC is equal to

A) \[2\sqrt{2}\]

B) \[3\sqrt{2}\]

C) \[2\sqrt{14}\]

D) \[2\sqrt{13}\]

Correct Answer: C

Solution :

Let  AD=3x, AG=2x, GD=x

DF=y, Given, AB=6, BC=4

\[BD=DDC=\frac{1}{2}BC=2\]

\[GD\times DF=BD\times CD\]

\[xy=4\] and \[A{{B}^{2}}=AG\times AF\]

\[36=2x\left( 3x+y \right)\]

\[36=6{{x}^{2}}+2xy\]\[\Rightarrow 6{{x}^{2}}=36-8\]\[\Rightarrow {{x}^{2}}=\frac{28}{6}\]

In \[\Delta ABC\], by Apollonius law,

\[A{{C}^{2}}+A{{B}^{2}}=2\left( A{{D}^{2}}+{{\left( \frac{BC}{2} \right)}^{2}} \right)\]\[\Rightarrow A{{C}^{2}}+36=2\left( 9{{x}^{2}}+4 \right)\]\[\Rightarrow A{{C}^{2}}+36=2\left( 9\times \frac{28}{6}+4 \right)\]\[\Rightarrow A{{C}^{2}}=2\left( 42+4 \right)-36\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,A{{C}^{2}}=56\]\[\Rightarrow AE=\sqrt{56}=2\sqrt{14}\]