JEE Main 2003 — Coordinate Geometry

Locus of a centriod of the triangle whose vertices are $(a \cos t, a \sin t),(b \sin t,-b \cos t)$ and $(1,0)$, where $t$ is a parameter, is

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then the value of $b^2$ is

The lines $2 \mathrm{x}-3 \mathrm{y}=5$ and $3 \mathrm{x}-4 \mathrm{y}=7$ are diameters of a circle having area as 154 sq. units. Then the equation of the circle is

The vectors $\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}} \& \overrightarrow{\mathrm{AC}}=5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$ are the sides of a triangle $\mathrm{ABC}$. The length of the median through $\mathrm{A}$ is