Question

Locus of the center of the circle which always passes through the fixed point $\left( {a,0} \right)$ and $\left( { - a,0} \right)$, where $a \ne 0$, is
A. $x = 1$
B. $x + y = 6$
C. $x + y = 2a$
D. $x = 0$

Answer 1

Hint: In this question, we are given the locus of the center of the circle passes through the point $\left( {a,0} \right)$ and $\left( { - a,0} \right)$. Let, the locus of the center of the circle be $\left( {h,k} \right)$. Now, take the square of distance of the locus from one point equal to the square of distance locus to the second point. Solve further using distance formula.

Formula Used:
Distance formula –
$d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $ where $\left( {{x_1},{y_1}} \right)$, $\left( {{x_2},{y_2}} \right)$ are the coordinates of the line.

Complete step by step solution:
Let, the coordinates of the locus of the center be $O\left( {h,k} \right)$
Also, let the given fixed points be $A\left( {a,0} \right)$and $B\left( { - a,0} \right)$,
Now, given that
Locus of the center of the circle passes through the point $A$ and $B$,
It implies that,
$A{O^2} = B{O^2}$
${\left( {\sqrt {{{\left( {h - a} \right)}^2} + {{\left( {k - 0} \right)}^2}} } \right)^2} = {\left( {\sqrt {{{\left( {h + a} \right)}^2} + {{\left( {k - 0} \right)}^2}} } \right)^2}$
${\left( {h - a} \right)^2} + {\left( {k - 0} \right)^2} = {\left( {h + a} \right)^2} + {\left( {k - 0} \right)^2}$
${h^2} + {a^2} - 2ha + {k^2} = {h^2} + {a^2} + 2ha + {k^2}$
$2ha + 2ha = 0$
$4ha = 0$, here $a \ne 0$
Therefore, $h = 0$
Here, the coordinate of the x-axis is zero.
$ \Rightarrow x = 0$

Option ‘D’ is correct

Note: In geometry, a locus is a set of points that satisfy a given condition or situation for a shape or figure. The locus is pluralized as loci. The region is the location of the loci. The term locus comes from the word location.