JEE Main & Advanced
JEE Main Paper Phase-I (Held on 07-1-2020 Evening)
question_answer
Let the tangents drawn from the origin to the circle, \[{{x}^{2}}+{{y}^{2}}8x4y+16=0\]touch it at the points A and B. The \[{{(AB)}^{2}}\]is equal to [JEE MAIN Held on 07-01-2020 Evening]
A)\[\frac{64}{5}\]
B)\[\frac{52}{5}\]
C)\[\frac{56}{5}\]
D)\[\frac{32}{5}\]
Correct Answer:
A
Solution :
[a] Equation of chord of contact is
\[x\cdot 0+y\cdot 0-4(x+0)-2(y+0)+16=0\] \[\therefore \,\,\,\,\,\,2x+y-8=0\] \[\therefore \] Length of \[CM=\left[ \frac{2\cdot 4+2-8}{\sqrt{{{2}^{2}}+{{1}^{2}}}} \right]=\frac{2}{\sqrt{5}}\] units. \[\therefore \,\,\,AM=BM=\sqrt{4-\frac{4}{5}}=\sqrt{\frac{16}{5}}\] \[\therefore \] Length of chord of contact (AB) \[=\frac{8}{\sqrt{5}}\] \[\therefore \] Square of length of chord of Contact\[={{\left( \frac{8}{\sqrt{5}} \right)}^{2}}=\frac{64}{5}\].