Question

The foot of the normal from the point $\left(\right.4,3\left.\right)$ to a circle is $\left(\right.2,1\left.\right)$ and a diameter of the circle has the equation $2x-y-2=0$ , then the equation of the circle is

• A. $x^{2} + y^{2} - 4 y + 2 = 0$
• B. $x^{2} + y^{2} - 4 y + 1 = 0$
• C. $x^{2} + y^{2} - 2 x - 1 = 0$
• D. $x^{2} + y^{2} - 2 x + 1 = 0$

Answer 1

Answer: (C)

Equation of the diameter of the circle is given as
$ \, 2x-y-2=0 \, $ ...(i)
If $P\left(\right.4,3\left.\right)$ and $N\left(\right.2,1\left.\right)$ are the given points, then
slope of $PN=\frac{3 - 1}{4 - 2}=1$
Equation of normal through $PN$ is


$y-1=\left(\right.x-2\left.\right)$
$x-y-1=0 \, \, \, $ ...(ii)
solving (i) and (ii), we get, the centre as $\left(\right.1, \, 0\left.\right)$
Hence, the equation of the circle is
$(x-1)^{2}+y^{2}=(2 - 1)^{2} \, + \, 1$
$x^{2}+y^{2}-2x-1=0$