Question

A foot of the normal from the point $(4,3)$ to a circle is $(2,1)$, and a diameter of the circle has the equation $2x-y=2$. Then the equation of the circle is

• A. $x^2+y^2+2x-1=0$
• B. $x^2+y^2-2x-1=0$
• C. $x^2+y^2-2x+1=0$
• D. none of these

Answer 1

Answer: (B)

$x^2+y^2-2x-1=0$

The equation of the line joining $(4,3)$ and $(2,1)$ is $L:x-y=1$
The point of intersection of L with the diameter gives the centre of the circle.
Intersection of $x-y=1$ and $2x-y=2is(x,y)=(1,0)$
The other end of the diagonal L be $(h,k)$.Centre$(1,0)$ is the mid point of $(2,1)$ and $(h,k)$
$\Rightarrow (\frac{h+2}{2},\frac{k+1}{2})=(1,0)$
$\Rightarrow$ h=0 and $k=-1$
The equation of the circle with $(0,-1)$ and $(2,1)$ as the ends of the diagonal is $\left(x\right)(x-2)+(y+1)(y-1)=0$