Question

The equation \[(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0\] represents a circle whose centre is
A $\left( {\dfrac{{{x_1} - {x_2}}}{2},\dfrac{{{y_1} - {y_2}}}{2}} \right)$
B $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
C $\left( {{x_1},{y_1}} \right)$
D $\left( {{x_2},{y_2}} \right)$

Answer 1

Hint: First, we will simplify the given equation \[(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0\] by multiplying the given factors. Then will use completing the square method to form the equation of the circle so that we can compare that with the general equation of the circle to find the center of the circle.

Complete step by step Solution:
 Given, the equation of circle \[(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0\].
\[(x - (x_1))(x - (x_2)) + (y - (y_1))(y - (y_2)) = 0\]
On multiplying the given factors of the above equation to get a simplified equation
${x^2} - x{x_2} - x{x_1} + {x_1}{x _2} + {y^2} - {y_1}y - {y_2}y + {y_1}{y_2} = 0$
Taking a common variable from the above equation
${x^2} - ({x_2} + {x_1})x + {x_1}{x _2} + {y^2} - ({y_1} + {y_2})y + {y_1}{y_2} = 0$
Using the completing the square method to get the required equation
${x^2} - ({x_2} + {x_1})x + {x_1}{x _2} + {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)^2} + {y^2} - ({y_1} + {y_2})y + {y_1}{y_2} + {\left( {\dfrac{{{y_1} + {y_2}}}{2}} \right)^2} = {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)^2} + {\left( {\dfrac{{{y_1} + {y_2}}}{2}} \right)^2}$
Collecting like terms such that we can make a square
\[{(x - \left( {\dfrac{{{x_1} + {x_2}}}{2}} \right))^2} + {x_1}{x_2} + {(y - \left( {\dfrac{{{y_1} + {y_2}}}{2}} \right))^2} + {y_1}{y_2} = {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)^2} + {\left( {\dfrac{{{y_1} + {y_2}}}{2}} \right)^2}\]
Shifting ${x_1}{x_2}$ and ${y_1}{y_2}$ to the other side
\[{(x - \left( {\dfrac{{{x_1} + {x_2}}}{2}} \right))^2} + {(y - \left( {\dfrac{{{y_1} + {y_2}}}{2}} \right))^2} = {\left( {\dfrac{{{x_1} + {x_2}}}{2}} \right)^2} + {x_1}{x_2} + {\left( {\dfrac{{{y_1} + {y_2}}}{2}} \right)^2} + {y_1}{y_2}\]
Comparing with standard equation of circle ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}$
Where $({x_1},{y_1})$ is the centre of the circle
In comparing we observe that the center of the given circle is
$\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$

Therefore, the correct option is B.

Additional Information: Given the circle's center and radius, the equation of a circle offers an algebraic approach to describing a circle. The formulas used to determine a circle's area or circumference are different from the equation for a circle. Numerous coordinate geometry problems involving circles employ this equation. The group of points whose separation from a fixed point has a constant value is represented by a circle. The radius of the circle, abbreviated r, is a constant that describes this fixed point, which is known as the circle's center. The standard equation of circle is ${\left( {x - {x_1}} \right)^2} + {\left( {y - {y_1}} \right)^2} = {r^2}$.

Note: Students should do calculations carefully to avoid any mistakes. They should avoid any direct calculations to avoid any errors and get correct solutions. They should use the given concept correctly.