Question

Line x = y touches a circle at point (1, 1). If the circle also passes through the point
(1, -3) then find its radius.
\[\begin{align}
  & a)3\sqrt{2} \\
 & b)3 \\
 & c)2\sqrt{2} \\
 & d)2 \\
\end{align}\]

Answer 1

Hint: Now we know that the equation of circle passing touching the line L = 0 at point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}+\lambda L=0$ hence we can write equation of circle in terms of $\lambda $ passing touching line x = y at point (1, 1). Now we know the circle passes through point (1, -3). Hence we will substitute the point in the equation to find the value of $\lambda $ . Now we have the equation of circle. We can easily find the radius of this circle.

Complete step by step answer:
We have that the line x = y touches the circle at point (1, 1).
We can write the equation x = y as x – y =0.
Now we know that the equation of circle passing touching the line L = 0 at point $\left( {{x}_{1}},{{y}_{1}} \right)$ is nothing but ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}+\lambda L=0$.
Hence the equation of circle touching the line x – y = 0 at point (1, 1) is
${{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}+\lambda \left( x-y \right)=0...................\left( 1 \right)$
Now we are given that the point (1, -3) lies on the circle.
Hence the point will satisfy the equation of circle. Now substituting x = 1 and y = - 3 in equation (1) we get.
$\begin{align}
  & {{\left( 1-1 \right)}^{2}}+{{\left( -3-1 \right)}^{2}}+\lambda \left( 1-\left( -3 \right) \right)=0 \\
 & \Rightarrow {{\left( -4 \right)}^{2}}+\lambda \left( 1+3 \right)=0 \\
 & \Rightarrow 16+4\lambda =0 \\
 & \Rightarrow 4\lambda =-16 \\
\end{align}$
Dividing the equation by 4 we get $\lambda =-4$
Now substituting this value in equation (1) we get
${{\left( x-1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}-4\left( x-y \right)=0$
Now we know that ${{\left( a\pm b \right)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab$
Hence we get
$\begin{align}
  & \left( {{x}^{2}}+1-2x \right)+\left( {{y}^{2}}+1-2y \right)-4x+4y=0 \\
 & \Rightarrow {{x}^{2}}-2x-4x+{{y}^{2}}-2y+4y+1+1=0 \\
 & \Rightarrow {{x}^{2}}-6x+{{y}^{2}}+2y+2=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-6x+2y+2=0 \\
\end{align}$
Now in the equation adding and subtracting 9 we get
\[\begin{align}
  & {{x}^{2}}-6x+9+{{y}^{2}}+2y+1+1-9=0 \\
 & \Rightarrow {{x}^{2}}-2\left( 3x \right)+{{\left( 3 \right)}^{2}}+{{y}^{2}}+2y+1=8 \\
\end{align}\]
Now again using the formula ${{\left( a\pm b \right)}^{2}}={{a}^{2}}+{{b}^{2}}\pm 2ab$ we get
${{\left( x-3 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=8$
Now the above equation is in the form \[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}\]
And the equation \[{{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}\] is nothing but equation of circle with center $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius r
Hence we get the radius of ${{\left( x-3 \right)}^{2}}+{{\left( y+1 \right)}^{2}}=8$ is $\sqrt{8}$
Now we know that $\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2}$ .
Hence the radius of the circle is $2\sqrt{2}$ units.
Option c is the correct option.

Note:
Now while using the equation ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}+\lambda L=0$ note that L should be of the form L = 0. If our line is not in this form then first write it in this form. Also we can find the radius directly as the radius of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$.