Question

A variable circle passes through the fixed point $\left( {2,0} \right)$ and touches the $y - $axis. What is the locus of the center of the circle?
A. A parabola
B. A circle
C. An ellipse
D. A hyperbola

Answer 1

Hint: First we will assume the coordinate of the center. Then find the distance of the center from the point $\left( {2,0} \right)$ and the distance between $y - $axis and the center. Equate the distances and replace $h$ by $x$ and $k$ by $y$to get the locus of the center. Then identify the equation.

Formula Used:
The distance between the point $\left( {{x_1},{y_1}} \right)$ and the line $ax + by + c = 0$ is $d = \left| {\dfrac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|$.
The distance between the points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.
The equation of $y - $axis is $x = 0$.

Complete step by step solution:
Assume that the center of the circle is $\left( {h,k} \right)$.
Calculate the distance between the points $\left( {h,k} \right)$ and $\left( {2,0} \right)$:
The distance between the points $\left( {h,k} \right)$ and $\left( {2,0} \right)$ is $\sqrt {{{\left( {h - 2} \right)}^2} + {{\left( {k - 0} \right)}^2}} $ unit.
The equation of $y - $axis is $x = 0$.
Calculate the distance between the points $\left( {h,k} \right)$ and $y - $axis.
Apply the distance formula between the point and line:
The distance between the points $\left( {h,k} \right)$ and $y - $axis is $\left| {\dfrac{h}{{{1^2}}}} \right| = h$units.
Equate the distances
$\sqrt {{{\left( {h - 2} \right)}^2} + {{\left( {k - 0} \right)}^2}} = h$
Take square both sides
$ \Rightarrow {\left( {\sqrt {{{\left( {h - 2} \right)}^2} + {{\left( {k - 0} \right)}^2}} } \right)^2} = {h^2}$
$ \Rightarrow {\left( {h - 2} \right)^2} + {\left( {k - 0} \right)^2} = {h^2}$
$ \Rightarrow {h^2} - 4h + 4 + {k^2} = {h^2}$
Subtract ${h^2}$ from both sides
$ \Rightarrow {h^2} - 4h + 4 + {k^2} - {h^2} = {h^2} - {h^2}$
$ \Rightarrow {k^2} - 4h + 4 = 0$
Replace $h$ by $x$ and $k$ by $y$
$ \Rightarrow {y^2} - 4x + 4 = 0$
$ \Rightarrow {y^2} = 4\left( {x - 1} \right)$
It is an equation of a parabola.

Option ‘A’ is correct

Note: We can solve the equation by using another method. Since the circle touches $y - $axis. So, the radius of the circle is the $x$ coordinate of the centre. If the centre of the circle is $\left( {h,k} \right)$, then the equation of the circle is ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {h^2}$. Since the circle passes through the point $\left( {2,0} \right)$. So substitute $\left( {2,0} \right)$ in the equation ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {h^2}$. To get the locus replace $h$ by $x$ and $k$ by $y$.