Question

find the equation of circle the coordinates of end points of whose diameter are (a, b) and (c, d) ​

Answer 1

The equation of required circle is:$\left (x- \dfrac{a+c}{2}\right )^2 + \left (y- \dfrac{b + d}{2} \right )^2 =\left ( \dfrac{a-c}{2} \right )^2 + \left ( \dfrac{b-d}{2} \right )^2$

Step-by-step explanation:

We know that;

• The equation of circle :

                                        $\left ( x-a \right )^2+\left ( y-b \right )^2 = r^2$

Here, (a, b) is the coordinate of center and 'r' is the radius.

• We are given the coordinates of end points of the diameter as (a,b) & (c,d).

• From these coordinates, we can find out the coordinates of center.

Coordinates of center = $\left ( \dfrac{a+c}{2},\dfrac{b+d}{2} \right )$

Also, we can find out the radius.

• $r = \sqrt{\left ( \dfrac{a+c}{2} - a \right )^2 + \left ( \dfrac{b+d}{2} - b \right )^2}$

• $r^2 = \left ( \dfrac{a+c}{2} - a \right )^2 + \left ( \dfrac{b+d}{2} - b \right )^2$

Now, we can find out the equation of circle as:

$\left (x- \dfrac{a+c}{2}\right )^2 + \left (y- \dfrac{b + d}{2} \right )^2 =\left ( \dfrac{a+c}{2} - a \right )^2 + \left ( \dfrac{b+d}{2} - b \right )^2$

$\left (x- \dfrac{a+c}{2}\right )^2 + \left (y- \dfrac{b + d}{2} \right )^2 =\left ( \dfrac{a-c}{2} \right )^2 + \left ( \dfrac{b-d}{2} \right )^2$

This is the equation of required circle.