Question

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

• A. $2\sqrt{3}$
• B. $3\sqrt{2}$
• C. $2\sqrt{2}$
• D. $\sqrt{2}$
• E. $2\sqrt{5}$

Answer 1

Answer: (E)

Given equation of parabola, $y=x^2-4x+3$
$\Rightarrow$ $(y+1)=x^2-4x+4$
$\Rightarrow$ ${(x-2)}^2=(y+1)$ Comparing with, $X^2=4aY$ here $X=x-2$ and $y=y+1,a=\frac{1}{4}$ Vertex is $(2,-1)=V$ (say) equation of circle, $x^2=9-{(y-3)}^2$ or ${(x-0)}^2+{(y-3)}^2=9$ Centre $C=(0,3)$ Now, $VC=\sqrt{{(2-0)}^2+{(-1-3)}^2}$
$=\sqrt{4+16}=\sqrt{20}$
$=2\sqrt{5}$