Question

Find the coordinates of a point A, where AB is the diameter of a circle whose centre is $ \left( {2, - 3} \right) $ and B is point $ \left( {1,4} \right) $ .

Answer 1

Hint: In order to determine the coordinate of the point A of the diameter AB ,Let the coordinate of Point A and B be $ A({x_1},{y_1}) $ and $ B({x_2},{y_2}) $ respectively. Use the midpoint formula on the diameter AB which is equal to $ M\left( {{m_1},{m_2}} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right) $ . The midpoint of any diameter is nothing but the centre of the circle which is given as $ \left( {2, - 3} \right) $ and the and the coordinate of point B is given as $ \left( {1,4} \right) $ . Put all these values in the midpoint formula and then Now comparing the x-coordinate of LHS with the x coordinate of RHS and similarly the y-coordinate of LHS with the y coordinate of RHS and simplify to get the coordinate of point $ A\left( {{x_1},{y_1}} \right) $ .

Complete step-by-step answer:
We are given a Circle $ C $ having its centre at coordinate $ \left( {2, - 3} \right) $ and a diameter AB whose coordinate of second end point B is $ \left( {1,4} \right) $ .
Since we have given diameter as $ AB $ and let the coordinates of the end points be $ A({x_1},{y_1}) $ and $ B({x_2},{y_2}) $ the coordinate of midpoint be $ M $ $ \left( {{m_1},{m_2}} \right) $ .
As we know the mid-point of every diameter is nothing but the centre of the circle and in our case the centre of circle is at $ \left( {2, - 3} \right) $ .Hence $ M\left( {{m_1},{m_2}} \right) = \left( {2, - 3} \right) $
 $ B({x_2},{y_2}) $ is given in our question as $ \left( {1,4} \right) $ .
We can find the midpoint of the diameter $ AB $ by using the coordinates of its endpoints.
We calculate the average of the x-coordinates and the average of the y-coordinates of the endpoints.
The formula of Midpoint: $ M\left( {{m_1},{m_2}} \right) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right) $
Putting the value of Point $ B({x_2},{y_2}) $ and the midpoint , we get
 $ \left( {2, - 3} \right) = \left( {\dfrac{{{x_1} + 1}}{2},\dfrac{{{y_1} + 4}}{2}} \right) $
Now comparing the x-coordinate of left-hand side with the x coordinate of right-hand side and similarly the y-coordinate of left-hand side with the y coordinate of right-hand side, we get
 $
  2 = \dfrac{{{x_1} + 1}}{2} \\
  4 = {x_1} + 1 \\
  {x_1} = 3 \;
 $
 $
  - 3 = \dfrac{{{y_1} + 4}}{2} \\
  - 6 = {y_1} + 4 \\
  {y_1} = - 10 \;
  $
 $ \therefore A\left( {{x_1},{y_1}} \right) = \left( {3, - 10} \right) $
Therefore, the coordinate of point A of the diameter AB is equal to $ A\left( {{x_1},{y_1}} \right) = \left( {3, - 10} \right) $ .
So, the correct answer is “(3, - 10)”.

Note: General equation of Circle having centre at $ \left( {h,k} \right) $ , is
 $ {\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2} $ where $ r $ is the radius of the circle
1. Compare the coordinate of the midpoint with the RHS coordinate carefully as x with x and y with y while obtaining the answer.
2.Diametre is the line which joins two points on the circle and passes through the centre but the chord only joins two points of the circle.
3. If the centre of the circle is not given , then consider it as $ \left( {0,0} \right) $ .