Chapter 1, Problem 110RE

To determine

To interpret:The given equation $x^2+y^2-6x-10y+34=0$ and decide if it represents a circle, a point or in the last case if it does not represent any graph.

Answer to Problem 110RE

The equation $x^2+y^2-6x-10y+34=0$ represents a point $\left(3,5\right)$ .

Explanation of Solution

Given information:

The equation $x^2+y^2-6x-10y+34=0$

Formula used:

For a given circle with center $\left(h,k\right)$ and radius $r$ the equation is given as:

  ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$

This is referred to as the Standard form for the equation of a given circle.

Moreover, if the center of the circle is nothing but the origin that is $\left(0,0\right)$ then the equation reduces to:

  $x^2+y^2=r^2$

Completing Square method:

To apply this method on an equation of the form: $a{x}^2+bx$

Method- Make sure that the coefficient of $x^2=1$ and then, add and subtract the square of the half of coefficient of $x$ and then club the terms using formulae:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+2ab+b^2\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$

Calculation:

Consider the given equation:

  $x^2+y^2-6x-10y+34=0$

It can also be written as:

  $\left(x^2-6x\right)+\left(y^2-10y\right)=-34$

Now apply Completing Square Method on the two brackets:

  $\left(x^2-6x+{\left(\frac{-6}{2}\right)}^2-{\left(\frac{-6}{2}\right)}^2\right)+\left(y^2-10y+{\left(\frac{-10}{2}\right)}^2-{\left(\frac{-10}{2}\right)}^2\right)=-34$

Now shift the extra terms to the right hand side

  $\left(x^2-6x+{\left(\frac{-6}{2}\right)}^2\right)+\left(y^2-10y+{\left(\frac{-10}{2}\right)}^2\right)=-34+{\left(\frac{-6}{2}\right)}^2+{\left(\frac{-10}{2}\right)}^2$

Recall the identities:

  $\begin{array}{l}{\left(a+b\right)}^2=a^2+2ab+b^2\\ {\left(a-b\right)}^2=a^2-2ab+b^2\end{array}$

Use this to get:

  $\begin{array}{l}\left(x^2-6x+3^2\right)+\left(y^2-10y+5^2\right)=-34+3^2+5^2\\ {\left(x-3\right)}^2+{\left(y-5\right)}^2=0\end{array}$

Now sum of two positive quantities can be $0$ only when each of them is individually equal to $0$ .

Therefore we get:

  $\begin{array}{l}{\left(x-3\right)}^2=0\\ x-3=0\\ x=3\end{array}$

And

  $\begin{array}{l}{\left(y-5\right)}^2=0\\ y-5=0\\ y=5\end{array}$

Thus,the equation $x^2+y^2-6x-10y+34=0$ represents a point $\left(3,5\right)$ .