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Q.
The distance between the vertex of the parabola $y = x^{2} - 4x + 3$ and the centre of the circle $x^{2} = 9 - ( y - 3 )^{2}$ is
Conic Sections
Solution:
Here, $y = x^{2} - 4x + 3$
$\Rightarrow y + 1 = x^{2} -4x + 4$
$\Rightarrow y+ 1 = \left(x -2 \right)^{2}$
So, the vertex is $\left(2 ,-1 \right)$
And for the circle, $x^{2} + \left(y - 3\right)^{2} = 9$, the centre is $\left(0, 3\right)$
$\therefore $ Distance between vertex and centre
$=\sqrt{2^{2}+\left(-4\right)^{2}}$
$=2\sqrt{5}$