Question

The locus of a point which moves such that the sum of the squares of its distances from three vertices of a $\Delta ABC$ is constant, is a circle whose center is at the
A. centroid of $\Delta ABC$
B. circumference of $\Delta ABC$
C. Orthocentre of $\Delta ABC$
D. incentre of $\Delta ABC$

Answer 1

Hint: In this question, we are given that the sum of the square of distances from three vertices of a $\Delta ABC$ to the locus is constant. Using this condition starts the solution. Now use the distance formula and solve further. You’ll get an equation similar to the general equation of a circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$. Compare them and coordinates of the circle will be there. Now, calculate the centroid of the triangle.

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.
General equation of circle, ${x^2} + {y^2} + 2gx + 2fy + c = 0$
Coordinates of the center of circle are $\left( {h,k} \right)$ where $h = - g$ and $k = - f$

Complete step by step solution:
Let, the coordinate of the vertices of $\Delta ABC$ be $A\left( {{x_1},{y_1}} \right)$, $B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$.
Also, let the coordinates of the locus be $P\left( {x,y} \right)$
Image: Triangle
Now, Given that
The locus of a point moves in such a way that the sum of the squares of its distances from three vertices of a $\Delta ABC$ is constant
It implies that,
${\left( {AP} \right)^2} + {\left( {BP} \right)^2} + {\left( {CP} \right)^2} = c$, $c$ is the constant
Using distance formula,
${\left( {\sqrt {{{\left( {x - {x_1}} \right)}^2} + {{\left( {y - {y_1}} \right)}^2}} } \right)^2} + {\left( {\sqrt {{{\left( {x - {x_2}} \right)}^2} + {{\left( {y - {y_2}} \right)}^2}} } \right)^2} + {\left( {\sqrt {{{\left( {x - {x_3}} \right)}^2} + {{\left( {y - {y_3}} \right)}^2}} } \right)^2} = c$
${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} + {\left( {x - {x_2}} \right)^2} + {\left( {y - {y_2}} \right)^2} + {\left( {x - {x_3}} \right)^2} + {\left( {y - {y_3}} \right)^2} = c$
${x^2} + {x_1}^2 - 2x{x_1} + {y^2} + {y_1}^2 - 2y{y_1} + {x^2} + {x_2}^2 - 2x{x_2} + {y^2} + {y_2}^2 - 2y{y_2} + {x^2} + {x_3}^2 - 2x{x_3} + {y^2} + {y_3}^2 - 2y{y_3} = c$
$3{x^2} + 3{y^2} + 2\left( { - {x_1} - {x_2} - {x_3}} \right)x + 2\left( { - {y_1} - {y_2} - {y_3}} \right)y + {x_1}^2 + {x_2}^2 + {x_3}^2 + {y_1}^2 + {y_2}^2 + {y_3}^2 - c = 0$
Divide the above equation by $3$,
${x^2} + {y^2} + 2\left( {\dfrac{{ - {x_1} - {x_2} - {x_3}}}{3}} \right)x + 2\left( {\dfrac{{ - {y_1} - {y_2} - {y_3}}}{3}} \right)y + \left( {\dfrac{{{x_1}^2 + {x_2}^2 + {x_3}^2 + {y_1}^2 + {y_2}^2 + {y_3}^2 - c}}{3}} \right) = 0 - - - \left( 1 \right)$
Now, compare equation (1) with the general equation of circle ${x^2} + {y^2} + 2gx + 2fy + c = 0$
Therefore, $g = \left( {\dfrac{{ - {x_1} - {x_2} - {x_3}}}{3}} \right),f = \left( {\dfrac{{ - {y_1} - {y_2} - {y_3}}}{3}} \right)$
Here, $h = - g = \dfrac{{{x_1} + {x_2} + {x_3}}}{3}$
$k = - f = \dfrac{{{y_1} + {y_2} + {y_3}}}{3}$
It means, the coordinates of center of the circle $\left( {h,k} \right) = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right) - - - - \left( 2 \right)$
Also, the coordinates of centroid of given triangle are $\left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right) - - - - \left( 3 \right)$
From equation (2) and (3)
The center of the circle is at the centroid of the given triangle $ABC$.

Option ‘A’ 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.