Question

A foot of 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 circle is

• A. $x^2+y^2=1$
• B. $x^2+y^2-2x+1=0$
• C. $x^2+y^2-2x-1=0$
• D. $x^2+y^2=5$

Answer 1

The correct option is C $x^2+y^2-2x-1=0$

The foot of the normal from $(4,3)$ to a circle is $(2,1)$ then, the equation of the normal to circle is
$y-1=\frac{2}{2}(x-2)$
$y-x=-1$ -----(1)
A diameter of the circle is $2x-y=2$ -----(2)
$\therefore$ Point of intersection of (1) & (2) is centre of circle,
$x=1$ and $y=0$
So, the equation of circle is
$(x-1{)}^2+(y-0{)}^2=r^2$
$(2,1)$ lies on the circle
$\therefore (2-1{)}^2+(1-0{)}^2=r^2$
$\Rightarrow r^2=2$
$\Rightarrow r=\sqrt{2}$
Hence, the equation of the circle is
$x^2+y^2-2x+1=2$
$\Rightarrow x^2+y^2-2x-1=0$
Hence, option 'C' is correct.