Question

The equation of the circle in diameter form with centre $(4,–2)$ and passing through the point $(2,-2)$ is

• A. $(x-2)(x+6)+(y+3)(y-2)=0$
• B. $(x+4)(x+2)+(y-1)(y+2)=0$
• C. $(x-6)(x+2)+(y-4)(y+2)=0$
• D. $(x-6)(x-2)+(y+2)(y+2)=0$

Answer 1

The correct option is D $(x-6)(x-2)+(y+2)(y+2)=0$

Let $(2,-2)$ be one end of the diameter.
We know that centre is the midpoint of two extreme points of diameter.
Let the other end-point of diameter be $(h,k)$
$\therefore 4=\frac{2+h}{2}\Rightarrow h=6$
and $-2=\frac{-2+k}{2}\Rightarrow k=-2$
$\therefore (h,k)=(6,-2)$
Hence, equation of the circle in diameter form is
$(x-6)(x-2)+(y+2)(y+2)=0$