Question

Find the equation of the circle with end points of the diamter as $(2,-1)$ and $(3,2)$.

• A. $x^2+y^2-5x-y+4=0$
• B. $x^2+y^2+5x+y+4=0$
• C. $x^2+y^2+5x+y-4=0$
• D. $x^2+y^2-5x-y-4=0$

Answer 1

The correct option is A $x^2+y^2-5x-y+4=0$

The equation of circle with its end points of diameter $(x_1,y_1)$and$(x_2,y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$

Here, $(x_1,y_1)\equiv (2,-1)$ and $(x_2,y_2)\equiv (3,2)$

Thus, equation of circle :- $(x-2)(x-3)+(y-(-1\left)\right)(y-2)=0$

$\Rightarrow x^2-5x+6+y^2-y-2=0$

$\Rightarrow x^2+y^2-5x-y+4=0$