Question

Circles (a) Complete the square to determine whether the equation represents a circle or a point or has no graph. (b) If the equation is that of circle, find its center and radius, and sketch the graph.

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

Short answer

(a) The equation$x^2+y^2-6x-10y+34=0$ has no graph.

(b) There is no center and radius of the given equation.

Step-by-step solution

Part a. Step 1. Write the given information.

The given equation is,

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

To make$x^2+bx$ a perfect square, add ${\left(\frac{b}{2}\right)}^2$, the square of the half of the coefficient of x both side of the equation.

The formula for perfect square is,

$x^2+bx+{\left(\frac{b}{2}\right)}^2={\left(x+\frac{b}{2}\right)}^2$

Part a. Step 2. Interpret the given equation.

Consider the given equation.

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

Group the terms of the above equation.

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

There are two expressions that needs to be converted to perfect square.$\left(x^2-6x\right)$ and $\left(y^2-10y\right)$.

Part a. Step 3. Convert the equation to perfect square.

Substitute$-6$ for b from first term of above equation, and$-10$ for b for the second term of the above equation.

$$\begin{gathered} \left(x^2-6x+{\left(\frac{-6}{2}\right)}^2\right)+\left(y^2-10y+{\left(\frac{-10}{2}\right)}^2\right)=-34+{\left(\frac{-10}{2}\right)}^2+{\left(\frac{-6}{2}\right)}^2 \\ \left(x^2-6x+9\right)+\left(y^2-10y+25\right)=-34+9+25 \\ {\left(x-3\right)}^2+{\left(y-5\right)}^2=0\mathrm{...}\left(1\right) \end{gathered}$$

The standard equation of circle is given by,

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

Here,$\left(h,k\right)$ is the center the of the circle and r is the radius of the circle.

Compare equation (1) and (2).

$\begin{array}{c}h=3\\ k=5\\ r=0\end{array}$

The radius of the circle is zero so it does not represent a circle. So no graph is possible.

Thus, the given equation has no graph.

Part b. Step 1. Write the given information.

The given equation is,

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

To make$x^2+bx$ a perfect square, add ${\left(\frac{b}{2}\right)}^2$, the square of the half of the coefficient of x both side of the equation.

The formula for perfect square is,

$x^2+bx+{\left(\frac{b}{2}\right)}^2={\left(x+\frac{b}{2}\right)}^2$

Part b. Step 2. Interpret the given equation.

Consider the given equation.

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

Group the terms of the above equation.

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

There are two expressions that needs to be converted to perfect square.$\left(x^2-6x\right)$ and $\left(y^2-10y\right)$.

Part b. Step 3. Convert the equation to perfect square.

Substitute$-6$ for b from first term of above equation, and$-10$ for b for the second term of the above equation.

$$\begin{gathered} \left(x^2-6x+{\left(\frac{-6}{2}\right)}^2\right)+\left(y^2-10y+{\left(\frac{-10}{2}\right)}^2\right)=-34+{\left(\frac{-10}{2}\right)}^2+{\left(\frac{-6}{2}\right)}^2 \\ \left(x^2-6x+9\right)+\left(y^2-10y+25\right)=-34+9+25 \\ {\left(x-3\right)}^2+{\left(y-5\right)}^2=0\mathrm{...}\left(1\right) \end{gathered}$$

The standard equation of circle is given by,

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

Here,$\left(h,k\right)$ is the center the of the circle and r is the radius of the circle.

Compare equation (1) and (2).

$\begin{array}{c}h=3\\ k=5\\ r=0\end{array}$

The radius of the circle is zero so it does not represent a circle. So no graph is possible.

Since, the given equation does not represent a circle so, there is no center and radius.