Question:
Find the equation of the circle, the coordinates of the end points of one of whose diameters are
A(3, 2) and B(2, 5)
Solution:
The equation of a circle passing through the coordinates of the end points of diameters is:
$\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$
Substituting, values: $\left(x_{1}, y_{1}\right)=(3,2) \&\left(x_{2}, y_{2}\right)=(2,5)$
We get:
$(x-3)(x-2)+(y-2)(y-5)=0$
$\Rightarrow x^{2}-2 x-3 x+6+y^{2}-5 y-2 y+10=0$
$\Rightarrow x^{2}+y^{2}-5 x-7 y+16=0$
Ans: $x^{2}+y^{2}-5 x-7 y+16=0$