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. $x^2+y^2-2y-1=0$
• D. None of these

Answer 1

The correct option is B

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

The explanation for the correct answer:

Step 1: Find the equation of the line:

Given: The normal from the point $(x_1,y_1)$ is $(4,3)$ to a circle $(x_2,y_2)$is $(2,1)$ and

The diameter of the circle is $2x-y=2$

The line equation that joints these two points is $y-y_1=m(x-x_1)$

where m is the slope of the given line.

Slope, m = $\frac{y_2-y_1}{x_2-x_1}$

Thus the equation of the line is $y-3=\frac{1-3}{2-4}(x-2)$

$\Rightarrow$ $y-3=1(x-4)$

$\Rightarrow$ $x-y=1\dots \dots .\left[1\right]$

The diameter of the circle is $2x-y=2\dots \dots ..\left[2\right]$

The point of intersection of these lines gives the center of the circle.

By subtracting the equations [1] and [2], we get $x=1$and $y=0$

Thus the equation of the circle can be obtained from the center $(x_1,y_1)$ is $(1,0)$ and the point $(x_2,y_2)$is $(2,1)$.

Step 2: Find the equation of the circle

The radius of the circle is given by $r=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$=\sqrt{{(2-1)}^2+{(1-0)}^2}$

$=\sqrt{1+1}$

$=\sqrt{2}$

The equation of the circle is given as ${(x-x_1)}^2+{(y-y_1)}^2=r^2$

$\Rightarrow$ ${(x-1)}^2+{(y-0)}^2={\left(\sqrt{2}\right)}^2$

$\Rightarrow$ $x^2-2x+1+y^2=2$

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

Hence, option (B) is the correct option.