Question

The distance between the vertex of the parabola $y=x^2-4x+3$and the centre of the circle $x^2=9-{\left(y-3\right)}^2$ is

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

Answer 1

The correct option is E

$2\sqrt{5}$ units

Explanation for the correct option:

Step 1: Finding the vertex

Given the vertex of the parabola $y=x^2-4x+3$and the centre of the circle $x^2=9-{\left(y-3\right)}^2$

Simplifying the parabola equation, we get

$\begin{array}{rcl}y& =& x^2-4x+3\\ y& =& x^2-4x+4-1\\ (y+1)& =& {(x-2)}^2\\ X& =& x-2\\ Y& =& y+1\\ \text{Vertex}& =& \left(2,-1\right)\end{array}$

Step 2: Finding the centre

Rewriting the equation for the centre of the circle $x^2=9-{\left(y-3\right)}^2$

$$\begin{gathered} x^2+{\left(y-3\right)}^2=3^2 \\ {(x-a)}^2+{(y-b)}^2=r^2,where(a,b)arecenter \end{gathered}$$

Thus, the Centre of the circle is $\left(0,3\right)$

Step 3: Finding the distance

Finding the distance between the centre $\left(0,3\right)$ and the vertex $\left(2,-1\right)$ , we get

$\begin{array}{rcl}D& =& \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ & =& \sqrt{{\left(2-0\right)}^2+{\left(-1-3\right)}^2}\\ & =& \sqrt{4+16}\\ & =& \sqrt{20}\\ & =& 2\sqrt{5}\end{array}$

The distance between the centre and the vertex is $2\sqrt{5}$ units.

Hence, option (E) is the correct answer.