Chapter 12, Problem 2CLT

To determine

Identify the conic and sketch its graph.

Answer to Problem 2CLT

The conic is circle.

Explanation of Solution

Given information:

Identify the conic and sketch its graph.

  $x^2+y^2-2x-4y+1=0$

Calculation:

Consider the given equation,

  $x^2+y^2-2x-4y+1=0$

To identify conic type rearranges the equation in standard form,

  $x^2+y^2-2x-4y+1=0$

Standard form is.

  $A{x}^2+C{y}^2+Dx+Ey+F=0$

  $x^2+y^2-2x-4y+1=0$

On comparing with standard form the value of ,

  $\begin{array}{l}A=1\\ C=1\end{array}$

To identify the conic type consider these conditions,

  $\begin{array}{cc}\begin{array}{l}conic\\ type\end{array}& condition\\ Circle& A=C\\ Parabola& AC=0\\ Ellipse& AC>0\\ Hyperbola& AC<0\end{array}$

Now

  $\begin{array}{l}A=1\\ C=1\end{array}$

Both values are of same sign and greater than zero and of equal value, so the given equation represents as conic type is Circle.

  $x^2+y^2-2x-4y+1=0$

The standard form of circle equation,

  $\begin{array}{l}{\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2\\ here\\ x,h\end{array}$

Centre of the circle.

So,

  $x^2+y^2-2x-4y+1=0$

Add $4$ both sides of the above expression,

  $\begin{array}{l}\left(x^2-2x+1\right)+\left(y^2-4x+4\right)=4\\ {\left(x-1\right)}^2+{\left(y-2\right)}^2=4\end{array}$

Now draw map of equation, ${\left(x-1\right)}^2+{\left(y-2\right)}^2=4$

  

Hence the conic is circle.