Explanation:
B Given,

\(\mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}-8 \mathrm{y}+21=0\)
\(2 \mathrm{~g}=-6, 2 \mathrm{f}=-8, \mathrm{c}=21\)
\(\mathrm{~g}=-3, \mathrm{f}=-4 \mathrm{c}=21\)
\(\mathrm{r}=\sqrt{\mathrm{g}^2+\mathrm{f}^2-\mathrm{c}}=\sqrt{(-3)^2+(-4)^2-21}\)
\(=\sqrt{9+16-21}=\sqrt{4}=2\)
\(\mathrm{r}=2, \text { centre }(3,4)\)
\(\text { equation (i) can be written as }\)
\((\mathrm{x}-3)^2+(\mathrm{y}-4)^2=4\)
Clearly the line \(\mathrm{AB}\) is chord of contact and its equation is,
\(\mathrm{xx}_1+\mathrm{yy}_1+\mathrm{g}\left(\mathrm{x}+\mathrm{x}_1\right)+\mathrm{f}\left(\mathrm{y}+\mathrm{y}_1\right)+\mathrm{c}=0\)
Here \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=(0,0)\)
\(0+0-3(\mathrm{x}+0)-4(\mathrm{y}+0)+21=0\)
\(3 \mathrm{x}+4 \mathrm{y}-21=0\)
Perpendicular distance from \((3,4)\) to the line \(3 x+4 y-\)
\(21=0\)
\(\mathrm{CM}=\frac{3(3)+4(4)-21}{\sqrt{9+16}}=\frac{4}{5}\)
\(\mathrm{AM}=\sqrt{\mathrm{Ac}^2-\mathrm{CM}^2}=\sqrt{4-\frac{16}{25}}=\sqrt{\frac{84}{5}}=\frac{2 \sqrt{21}}{5}\)
\(\therefore \mathrm{AB}=2 \mathrm{AM}=2 \times \frac{2 \sqrt{21}}{5}=\frac{4 \sqrt{21}}{5}\)