Explanation:
E Given,
\(y=x^2-4 x+3\)
centre of the circle \(x^2=9-(y-3)^2\)
\(y=x^2-4 x+3\)
\(y=x^2-4 x+4-1\)
\((y+1)=(x-2)^2\)
\(A=(x-2)\)
\(B=y+1\)
Vertex of the parabola \((2,-1)\) centre of the circle
\(x^2=9-(y-3)^2\)
\(x^2+(y-3)^2=9\)
\(x^2+(y-3)^2=3^2\)
\((x-a)^2+(y-b)^2=r^2\)
Comparing equation (i) and (ii) we get \(\mathrm{a}=0, \mathrm{~b}=3\)
\(\because\) Centre of the circle \((0,3)\)
The distance between the \((0,3)\) centre and the vertex \((2\), \(-1)\)
\(\mathrm{D}=\sqrt{\left(\mathrm{x}_2-\mathrm{x}_1\right)^2+\left(\mathrm{y}_2-\mathrm{y}_1\right)^2}\)
\(\because\left[\begin{array}{l}\mathrm{x}_1=0, \mathrm{y}_1=3 \\ \mathrm{x}_2=2, \mathrm{y}_2=-1\end{array}\right]\)
\(=\sqrt{(2-0)^2+(-1-3)^2}\)
\(=\sqrt{4+16}\)
\(=\sqrt{20}\)
\(\mathrm{D}=2 \sqrt{5}\)
