The line $2x-y+1=0$ is tangent to the circle at the point $\left( 2,5 \right)$ and the centre of the circle lies on $x-2y=4$ . Then find the radius of the circle.
Answer
365.4k+ views
Hint: Sketch the diagram first to visualize the question better, then you could arrive at the answer merely by finding the perpendicular distance between the two lines.
Let’s sketch the situation given to us in the question. So we have a circle that has the line
$2x-y+1=0$ acting as a tangent to it at the point $(2,5)$, and we have the centre of the circle lying on the line $x-2y=4$. Drawing the figure while keeping all these features in mind, the diagram we’ll get is:
The lines do look parallel in the diagram, but from the equations of the lines, we can clearly see that they are not parallel.
Let’s assume the $y$ coordinate of the centre of the circle to be $=t$. Since we’re talking about the centre, the point will definitely satisfy the equation $x-2y=4$. Thus, we can find the $x$ coordinate of the centre simply by substituting for $y=t$ in the equation of the line. Doing so, we get :
$\begin{align}
& x-2y=4 \\
& \Rightarrow x=4+2y \\
& \Rightarrow x=4+2\left( t \right)=4+2t \\
\end{align}$
So, the coordinates of the centre are :
$\therefore c\equiv \left( 4+2t,t \right)$
Now, let’s find out the distance between the centre and the tangent line. That will be found out by applying the formula using which we can find the perpendicular distance of any point $(X,Y)$ from the line $Ax+By+C=0$. So, if we assume the perpendicular distance of point $(X,Y)$ from the line $Ax+By+C=0$ is $d$, then : $d=|\dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}|$
Thus, here our point $(X,Y)=(4+2t,t)$ and our line $Ax+By+C=0$ is equal to the line $2x-y+1=0$.
Therefore, the perpendicular distance $d$ of point $(4+2t,t)$ from line $2x-y+1=0$ is :
\[\begin{align}
& d=\left| \dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right| \\
& \Rightarrow d=\left| \dfrac{2\left( 4+2t \right)-\left( t \right)+1}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} \right| \\
& \Rightarrow d=\left| \dfrac{8+4t-t+1}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} \right| \\
& \Rightarrow d=\left| \dfrac{\left( 3t+9 \right)}{\sqrt{5}} \right|...............(1) \\
\end{align}\]
And since, from the figure, we can see that this distance will also be equal to the radius of the circle,
$d=r=|\dfrac{3t+9}{\sqrt{5}}|$
Now, since the tangent touches the circle at the point $(2,5)$, the distance of the centre from this point will also be equal to the radius. So, by distance formula $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$, we get :
\[\begin{align}
& r=\sqrt{{{\left( (4+2t)-2 \right)}^{2}}+{{\left( t-5 \right)}^{2}}} \\
& r=\sqrt{{{\left( 2+2t \right)}^{2}}+{{\left( t-5 \right)}^{2}}}............(\text{2)} \\
\end{align}\]
By equating $(1)$ and $(2)$ with each other, we get;
$|\dfrac{\left( 3t+9 \right)}{\sqrt{5}}|=\sqrt{{{\left( 2+2t \right)}^{2}}+{{\left( t-5 \right)}^{2}}}$
Squaring both sides, we get :
$\begin{align}
& \Rightarrow \dfrac{9{{t}^{2}}+81+54t}{5}={{(2+2t)}^{2}}+{{(t-5)}^{2}} \\
& \Rightarrow 9{{t}^{2}}+81+54t=5[4+4{{t}^{2}}+8t+{{t}^{2}}+25-10t] \\
\end{align}$
$\begin{align}
& \Rightarrow 9{{t}^{2}}-54t+81=5\left( 5{{t}^{2}}-2t+29 \right) \\
& \Rightarrow 16{{t}^{2}}-64t+64=0 \\
\end{align}$
Dividing by $16$ on both sides, we get :
$\Rightarrow {{t}^{2}}-4t+4=0$
By factorising , we get :
$\begin{align}
& {{\left( t-2 \right)}^{2}}=0 \\
& \therefore t-2=0 \\
& \\
\end{align}$
and the centre’s coordinates become :
$\begin{align}
& C\left( 4+2t,t \right) \\
& C\left( 4+\left( 2\times 2 \right),2 \right) \\
& \therefore \\
\end{align}$
Now the radius will become;
$r=\sqrt{5{{\left( 2 \right)}^{2}}-\left( 2 \right)\times \left( 2 \right)+29}$
\[\therefore r=\sqrt{45}=3\sqrt{5}\] units.
Therefore, the radius of the circle is $3\sqrt{5}$ units.
Note: Most of the student make mistake in remembering formulae of distance i.e. $d=\left| \dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$ and finding radius, avoid calculation mistakes.
Let’s sketch the situation given to us in the question. So we have a circle that has the line
$2x-y+1=0$ acting as a tangent to it at the point $(2,5)$, and we have the centre of the circle lying on the line $x-2y=4$. Drawing the figure while keeping all these features in mind, the diagram we’ll get is:

The lines do look parallel in the diagram, but from the equations of the lines, we can clearly see that they are not parallel.
Let’s assume the $y$ coordinate of the centre of the circle to be $=t$. Since we’re talking about the centre, the point will definitely satisfy the equation $x-2y=4$. Thus, we can find the $x$ coordinate of the centre simply by substituting for $y=t$ in the equation of the line. Doing so, we get :
$\begin{align}
& x-2y=4 \\
& \Rightarrow x=4+2y \\
& \Rightarrow x=4+2\left( t \right)=4+2t \\
\end{align}$
So, the coordinates of the centre are :
$\therefore c\equiv \left( 4+2t,t \right)$
Now, let’s find out the distance between the centre and the tangent line. That will be found out by applying the formula using which we can find the perpendicular distance of any point $(X,Y)$ from the line $Ax+By+C=0$. So, if we assume the perpendicular distance of point $(X,Y)$ from the line $Ax+By+C=0$ is $d$, then : $d=|\dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}}|$
Thus, here our point $(X,Y)=(4+2t,t)$ and our line $Ax+By+C=0$ is equal to the line $2x-y+1=0$.
Therefore, the perpendicular distance $d$ of point $(4+2t,t)$ from line $2x-y+1=0$ is :
\[\begin{align}
& d=\left| \dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right| \\
& \Rightarrow d=\left| \dfrac{2\left( 4+2t \right)-\left( t \right)+1}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} \right| \\
& \Rightarrow d=\left| \dfrac{8+4t-t+1}{\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 1 \right)}^{2}}}} \right| \\
& \Rightarrow d=\left| \dfrac{\left( 3t+9 \right)}{\sqrt{5}} \right|...............(1) \\
\end{align}\]
And since, from the figure, we can see that this distance will also be equal to the radius of the circle,
$d=r=|\dfrac{3t+9}{\sqrt{5}}|$
Now, since the tangent touches the circle at the point $(2,5)$, the distance of the centre from this point will also be equal to the radius. So, by distance formula $\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$, we get :
\[\begin{align}
& r=\sqrt{{{\left( (4+2t)-2 \right)}^{2}}+{{\left( t-5 \right)}^{2}}} \\
& r=\sqrt{{{\left( 2+2t \right)}^{2}}+{{\left( t-5 \right)}^{2}}}............(\text{2)} \\
\end{align}\]
By equating $(1)$ and $(2)$ with each other, we get;
$|\dfrac{\left( 3t+9 \right)}{\sqrt{5}}|=\sqrt{{{\left( 2+2t \right)}^{2}}+{{\left( t-5 \right)}^{2}}}$
Squaring both sides, we get :
$\begin{align}
& \Rightarrow \dfrac{9{{t}^{2}}+81+54t}{5}={{(2+2t)}^{2}}+{{(t-5)}^{2}} \\
& \Rightarrow 9{{t}^{2}}+81+54t=5[4+4{{t}^{2}}+8t+{{t}^{2}}+25-10t] \\
\end{align}$
$\begin{align}
& \Rightarrow 9{{t}^{2}}-54t+81=5\left( 5{{t}^{2}}-2t+29 \right) \\
& \Rightarrow 16{{t}^{2}}-64t+64=0 \\
\end{align}$
Dividing by $16$ on both sides, we get :
$\Rightarrow {{t}^{2}}-4t+4=0$
By factorising , we get :
$\begin{align}
& {{\left( t-2 \right)}^{2}}=0 \\
& \therefore t-2=0 \\
& \\
\end{align}$
and the centre’s coordinates become :
$\begin{align}
& C\left( 4+2t,t \right) \\
& C\left( 4+\left( 2\times 2 \right),2 \right) \\
& \therefore \\
\end{align}$
Now the radius will become;
$r=\sqrt{5{{\left( 2 \right)}^{2}}-\left( 2 \right)\times \left( 2 \right)+29}$
\[\therefore r=\sqrt{45}=3\sqrt{5}\] units.
Therefore, the radius of the circle is $3\sqrt{5}$ units.
Note: Most of the student make mistake in remembering formulae of distance i.e. $d=\left| \dfrac{AX+BY+C}{\sqrt{{{A}^{2}}+{{B}^{2}}}} \right|$ and finding radius, avoid calculation mistakes.
Last updated date: 30th Sep 2023
•
Total views: 365.4k
•
Views today: 3.65k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Who had given the title of Mahatma to Gandhi Ji A Bal class 10 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE
