Question

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

Hint: Any line which passes through origin cut the circumference perpendicularly. Equation of normal from any point passes through the center and the diameter also passes through the center. Find the equation of normal and solve it with the equation of diameter. The coordinate so obtained is the center of the circle. Find distance between center and point on circumference by using distance formula. This gives a radius of the circle, then applies the standard equation of the circle to find the equation of the circle.
Complete step by step solution:
As we know from the question that the foot of perpendicular from point (4, 3) on the circle is (2,1) so the line joining the two points is normal to the circle and it passes through origin. Here we have to find out the equation of line joining the point (4, 3) and (2,1).
Equation of line joining the two points$\left( {{x}_{1,}}{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ can be written as
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$
So equation of line joining the point (4,3) and (2,1)
$\Rightarrow y-3=\dfrac{1-3}{2-4}\left( x-4 \right)$ so we can write it as
$\begin{align}
  & \Rightarrow y-3=\dfrac{-2}{-2}\left( x-4 \right) \\
 & \Rightarrow y-3=\left( x-4 \right) \\
 & \Rightarrow y-x=-4+3 \\
 & \Rightarrow x-y=1---------(1) \\
\end{align}$
Now we have Equation of diameter is
$2x-y=2---------(2)$
Now subtracting equation (1) from equation (2)
$\begin{align}
  & \Rightarrow 2x-y-(x-y)=2-1 \\
 & \Rightarrow 2x-y-x+y=1 \\
 & \Rightarrow x=1 \\
\end{align}$
Now we put the value the of x in equation (1) so that we get value of y hence
$\begin{align}
  & \Rightarrow 1-y=1 \\
 & \Rightarrow y=0 \\
\end{align}$
So we get the coordinate of center of circle
$\therefore center\text{ of circle c}\equiv \text{(1,0)}$
Now radius of circle is distance between center of circle i.e (1,0) and point on circumference of circle i.e (2,1)
As we that distance between the two points$\left( {{x}_{1,}}{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ can be written as
$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
So radius of circle is
$\begin{align}
  & r=\sqrt{{{\left( 1-2 \right)}^{2}}+{{\left( 0-1 \right)}^{2}}} \\
 & \Rightarrow r=\sqrt{{{1}^{2}}+{{1}^{2}}} \\
 & \Rightarrow r=\sqrt{2} \\
\end{align}$
Now we know that standard equation of circle is
${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{\left( r \right)}^{2}}$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ is the center of circle and r is the radius of circle
So we can write ${{\left( x-1 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( \sqrt{2} \right)}^{2}}$
So on expanding we can write
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{1}^{2}}-2x+{{y}^{2}}=2 \\
 & \Rightarrow {{x}^{2}}-2x+{{y}^{2}}=2-1 \\
 & \Rightarrow {{x}^{2}}-2x+{{y}^{2}}=1 \\
 & \Rightarrow {{x}^{2}}-2x+{{y}^{2}}-1=0 \\
\end{align}$
Hence equation of circle is
${{x}^{2}}+{{y}^{2}}-2x-1=0$
So option B is correct.

Note: While solving the simultaneous equation we can solve it by elimination method also. By geometry normal are perpendicular to the tangent at the point on the circle and line joining center and point of contact at tangent point is perpendicular to the tangent so normal always passes through the center of the circle.