Question

Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose center is on the line \[x-3y-11=0\].

Answer 1

Hint: In the above question we will suppose a point \[\left( {{x}_{1}},{{y}_{1}} \right)\] center of the circle. Since this point lies on the line \[x-3y-11=0\]. So it satisfies by substituting \[\left( {{x}_{1}},{{y}_{1}} \right)\] in the equation of line, we get an equation between \[{{x}_{1}}\] and \[{{y}_{1}}\]. Also we know that the distance between the center of a circle to the point on the circle is constant. So we will use the concept to find out \[{{x}_{1}}\] and \[{{y}_{1}}\], then we will use the general formula for the equation of a circle as follows:
 \[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}\], where ‘r’ is the radius of the circle.

Complete step-by-step solution -
We have been given that a circle is passing through the points (2, 3) and (-1, 1) and whose center is on the line \[x-3y-11=0\].
So the point \[\left( {{x}_{1}},{{y}_{1}} \right)\] satisfied the equation of the line as follows:
\[\begin{align}
  & {{x}_{1}}-3{{y}_{1}}-11=0 \\
 & {{x}_{1}}-3{{y}_{1}}=11 \\
 & {{x}_{1}}=11+3{{y}_{1}}.....(1) \\
\end{align}\]
We know that the distance between the center of a circle and points on the circle is constant and equals to radius of the circle.
Also, the distance between two points, say, \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] is given by:
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
So the distance between the center \[\left( {{x}_{1}},{{y}_{1}} \right)\] and (2, 3) is given as follows:
\[{{d}_{1}}=\sqrt{{{\left( 2-{{x}_{1}} \right)}^{2}}+{{\left( 3-{{y}_{1}} \right)}^{2}}}....(2)\]
Also the distance between the center \[\left( {{x}_{1}},{{y}_{1}} \right)\] and (-1, 1) is given as follows:
\[{{d}_{2}}=\sqrt{{{\left( -1-{{x}_{1}} \right)}^{2}}+{{\left( 1-{{y}_{1}} \right)}^{2}}}....(3)\]
We know that both the distances \[{{d}_{1}}\] and \[{{d}_{2}}\]are equal as they are the radius of the circle.
So we will equate the equation (2) and (3).
\[\begin{align}
  & \Rightarrow {{d}_{1}}={{d}_{2}} \\
 & \Rightarrow \sqrt{{{\left( 2-{{x}_{1}} \right)}^{2}}+{{\left( 3-{{y}_{1}} \right)}^{2}}}=\sqrt{{{\left( -1-{{x}_{1}} \right)}^{2}}+{{\left( 1-{{y}_{1}} \right)}^{2}}} \\
\end{align}\]
On equating both the sides of the equation, we get as follows:
\[\begin{align}
  & \Rightarrow {{\left( \sqrt{{{\left( 2-{{x}_{1}} \right)}^{2}}+{{\left( 3-{{y}_{1}} \right)}^{2}}} \right)}^{2}}={{\left( \sqrt{{{\left( -1-{{x}_{1}} \right)}^{2}}+{{\left( 1-{{y}_{1}} \right)}^{2}}} \right)}^{2}} \\
 & ={{\left( 2-{{x}_{1}} \right)}^{2}}+{{\left( 3-{{y}_{1}} \right)}^{2}}={{\left( -1-{{x}_{1}} \right)}^{2}}+{{\left( 1-{{y}_{1}} \right)}^{2}} \\
\end{align}\]
We know that \[{{x}_{1}}=11+3y\], from equation (1). So by substituting the value of \[{{x}_{1}}\] in the above equation, we get as follows:
\[\begin{align}
  & \Rightarrow {{\left[ 2-\left( 11+3{{y}_{1}} \right) \right]}^{2}}+{{\left[ 3-{{y}_{1}} \right]}^{2}}={{\left[ -1-\left( 11+3{{y}_{1}} \right) \right]}^{2}}+{{\left( 1-{{y}_{1}} \right)}^{2}} \\
 & ={{\left( 2-11-3{{y}_{1}} \right)}^{2}}+{{\left( 3-{{y}_{1}} \right)}^{2}}={{\left( -1-11-3{{y}_{1}} \right)}^{2}}+{{\left( 1-{{y}_{1}} \right)}^{2}} \\
 & ={{\left( -9-3{{y}_{1}} \right)}^{2}}+{{\left( 3-{{y}_{1}} \right)}^{2}}=\left( -12-3{{y}_{1}} \right)+{{\left( 1-{{y}_{1}} \right)}^{2}} \\
\end{align}\]
On expanding each term of the equation, we get as follows:
\[\begin{align}
  & \Rightarrow {{\left( -9 \right)}^{2}}+{{\left( 3{{y}_{1}} \right)}^{2}}+2\left( -9 \right)\left( -3{{y}_{1}} \right)+{{\left( 3 \right)}^{2}}+{{\left( {{y}_{1}} \right)}^{2}}-2\times 3{{y}_{1}}={{\left( -12 \right)}^{2}}+{{\left( -3{{y}_{1}} \right)}^{2}}+2\left( -12 \right)\left( -3{{y}_{1}} \right)+{{1}^{2}}+{{y}_{1}}^{2}-2{{y}_{1}} \\
 & \Rightarrow 81+9{{y}_{1}}^{2}+54{{y}_{1}}+9+{{y}_{1}}^{2}-6{{y}_{1}}=144+9{{y}_{1}}^{2}+72{{y}_{1}}+1+{{y}_{1}}^{2}-2{{y}_{1}} \\
 & \Rightarrow 81+9+9{{y}_{1}}^{2}+{{y}_{1}}^{2}+54{{y}_{1}}-6{{y}_{1}}=144+1+9{{y}_{1}}^{2}+{{y}_{1}}^{2}+72{{y}_{1}}-2{{y}_{1}} \\
 & \Rightarrow 90+10{{y}_{1}}^{2}+48{{y}_{1}}=145+10{{y}_{1}}^{2}+70{{y}_{1}} \\
 & \Rightarrow 10{{y}_{1}}^{2}-10{{y}_{1}}^{2}=145-90+70{{y}_{1}}-48{{y}_{1}} \\
 & \Rightarrow 0=55+22{{y}_{1}} \\
 & \Rightarrow -55=22{{y}_{1}} \\
 & \Rightarrow \dfrac{-55}{22}={{y}_{1}} \\
 & \Rightarrow {{y}_{1}}=\dfrac{-5}{2} \\
\end{align}\]
Now substituting the value of \[{{y}_{1}}=\dfrac{-5}{2}\]in the equation (1), we get as follows:
\[\begin{align}
  & {{x}_{1}}=11+3{{y}_{1}} \\
 & {{x}_{1}}=11+3\left( \dfrac{-5}{2} \right) \\
 & {{x}_{1}}=11-\dfrac{15}{2} \\
 & {{x}_{1}}=\dfrac{22-15}{2} \\
 & {{x}_{1}}=\dfrac{7}{2} \\
\end{align}\]
Hence, the center of the circle is \[\left( \dfrac{7}{2},\dfrac{-5}{2} \right)\].
Now substituting the value of \[{{x}_{1}}\] and \[{{y}_{1}}\] in the equation (2) we will get the radius of the circle as follows:
Radius (r),
\[\begin{align}
  & =\sqrt{{{\left( 2-{{x}_{1}} \right)}^{2}}+{{\left( 3-{{y}_{1}} \right)}^{2}}} \\
 & =\sqrt{{{\left( 2-\dfrac{7}{2} \right)}^{2}}+{{\left( 3-\left( \dfrac{-5}{2} \right) \right)}^{2}}} \\
 & =\sqrt{{{\left( \dfrac{4-7}{2} \right)}^{2}}+{{\left( \dfrac{6+5}{2} \right)}^{2}}} \\
 & =\sqrt{{{\left( \dfrac{-3}{2} \right)}^{2}}+{{\left( \dfrac{11}{2} \right)}^{2}}} \\
 & =\sqrt{\dfrac{9}{4}+\dfrac{121}{4}}=\sqrt{\dfrac{130}{4}}=\dfrac{\sqrt{130}}{2} \\
\end{align}\]
So we get the radius of the circle as \[\dfrac{\sqrt{130}}{2}\] units.
We know that the equation of circle is given by:
\[\begin{align}
  & {{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}} \\
 & {{\left( x-\dfrac{7}{2} \right)}^{2}}+{{\left[ y-\left( \dfrac{-5}{2} \right) \right]}^{2}}={{\left( \dfrac{\sqrt{130}}{2} \right)}^{2}} \\
 & {{x}^{2}}+{{\left( \dfrac{7}{2} \right)}^{2}}-2x\left( \dfrac{7}{2} \right)+{{y}^{2}}+\dfrac{25}{4}+2y\left( \dfrac{5}{2} \right)=\dfrac{130}{4} \\
 & {{x}^{2}}+{{y}^{2}}-7x+5y+\dfrac{49}{4}+\dfrac{25}{4}=\dfrac{65}{2} \\
 & {{x}^{2}}+{{y}^{2}}-7x+5y+\dfrac{74}{4}=\dfrac{65}{2} \\
 & {{x}^{2}}+{{y}^{2}}-7x+5y+\dfrac{37}{2}=\dfrac{65}{2} \\
 & {{x}^{2}}+{{y}^{2}}-7x+54=\dfrac{65}{2}-\dfrac{37}{2} \\
 & {{x}^{2}}+{{y}^{2}}-7x+54=\dfrac{65-37}{2} \\
 & {{x}^{2}}+{{y}^{2}}-7x+54=\dfrac{28}{2} \\
 & {{x}^{2}}+{{y}^{2}}-7x+54=14 \\
\end{align}\]
Therefore, the required equation of the circle is equal to \[{{x}^{2}}+{{y}^{2}}-7x+54=14\]. We can represent it as below,

Note: Be careful while squaring and equating the equation (2) and (3) as there is a chance of calculation mistake. Also, take care of the signs when you substitute \[x-3y-11=0\] in the equation. Here key concept to solve this type of question is that every point on the circle is always equidistant from the centre of the circle.