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JEE MAIN 2025
23-01-2025 SHIFT-1
Question
Let $\left| {\frac{{\bar z - i}}{{2\bar z + i}}} \right| = \frac{1}{3},z \in C$ , be the equation of a circle with center at C. If the area of the triangle, whose vertices are at the points $(0,0),{\rm{C}}$ and $(\alpha ,0)$ is 11 square units, then ${\alpha ^2}$ equals:
Select the correct option:
A
$\frac{{121}}{{25}}$
B
100
C
$\frac{{81}}{{25}}$
D
50
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\left| {\frac{{\bar z - i}}{{2\bar z + i}}} \right| = \frac{1}{3}$
$\left| {\frac{{\bar z - i}}{{\bar z + \frac{i}{2}}}} \right| = \frac{2}{3}$
$3|x - iy - i| = 2\left| {x - iy + \frac{i}{2}} \right|$
$9\left( {{x^2} + {{(y + 1)}^2}} \right) = 4\left( {{x^2} + {{(y - 1/3)}^2}} \right)$
$9{x^2} + 9{y^2} + 18y + 9 = 4{x^2} + 4{y^2} - 4y + 1$
$5{x^2} + 5{y^2} + 22y + 8 = 0$
${x^2} + {y^2} + \frac{{22}}{5}y + \frac{8}{5} = 0$
${\rm{ centre }} \Rightarrow \left( {0, - \frac{{11}}{5}} \right)$
$\left| {\frac{1}{2}\left| {\begin{array}{*{20}{c}} 0&0&1\\ 0&{ - 11/5}&1\\ \alpha &0&1 \end{array}} \right|} \right|{\rm{ }}{\mkern 1mu} = 11$
$\Rightarrow {\left( { - \frac{{11}}{5}\alpha } \right)^2} = {(11 \times 2)^2}$
$ \Rightarrow {\alpha ^2} = 100$
Question Tags
JEE Main
Mathematics
Medium
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