Question

If one end of the diameter of the circle ${x^2} + {y^2} - 8x - 4y + c = 0$ is $\left( { - 3,2} \right)$then other coordinate is
A. $\left( {5,3} \right)$
B. $\left( {6,2} \right)$
C. $\left( {1, - 8} \right)$
D. $\left( {11,2} \right)$

Answer 1

Hint: In this question, we are given a equation of circle ${x^2} + {y^2} - 8x - 4y + c = 0$ and the coordinate of one end of its diameter and we have to find the other end. First step is to calculate the center points by comparing the given equation with the standard form of the equation of circle. Then, apply the mid-term formula.

Formula Used:
The general form of the equation of circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ whose center is $\left( { - g, - f} \right)$
Midpoint formula $\left( {x = \dfrac{{{x_1} + {x_2}}}{2},y = \dfrac{{{y_1} + {y_2}}}{2}} \right)$ where $\left( {x,y} \right)$ is the coordinate of midpoint and $\left( {{x_1},{y_1}} \right)$, $\left( {{x_2},{y_2}} \right)$ are the coordinates of the end points of the line.

Complete step by step solution:
Given that,
The coordinate of one end of the diameter of the circle ${x^2} + {y^2} - 8x - 4y + c = 0$ is $\left( { - 3,2} \right)$.

Image: Circle
Compare the above equation with ${x^2} + {y^2} + 2gx + 2fy + c = 0$whose center is $\left( { - g, - f} \right)$
It implies that, the center of the circle is at $\left( {4,2} \right)$
Now, the coordinate of the diameter of one end is $\left( { - 3,2} \right)$
Let, the coordinates of other end be $\left( {{x_1},{y_1}} \right)$
Using midpoint formula,
$\dfrac{{{x_1} - 3}}{2} = 4,\dfrac{{{y_1} + 2}}{2} = 0$
$ \Rightarrow {x_1} = 11,{y_1} = 2$
The coordinate of the other point are $\left( {11,2} \right)$

Option ‘D’ is correct

Note: To solve such a question, first try to make the figure and understand the question properly. One should always remember each and every equation of the circle (General and standard both). Also, if the line is divided in two equal halves and any two coordinates are given, always apply a mid-term formula to find the third coordinate.