# How do you decide whether or not the equation below has a circle as its graph If it does, give the center and the radius. If it does not, describe the graph 25x^2 +25y^2- 30x+ 30y -18 =0?

Jan 14, 2017

The eqn. represents a circle having centre $\left(\frac{3}{5} , - \frac{3}{5}\right)$ & radius

$\frac{6}{5}$.

#### Explanation:

The General Second Degree Equation in ${\mathbb{R}}^{2}$

$a {x}^{2} + 2 h x y + b {y}^{2} + 2 g x + 2 f y + c = 0$ will represent a Circle , if,

(i): a=b!=0, (ii): h=o, &, (iii): g^2+f^2-ac>0.

In the event, its Centre is $\left(- \frac{g}{a} , - \frac{f}{a}\right)$ and Radius is

$\frac{\sqrt{{g}^{2} + {f}^{2} - a c}}{|} a |$.

In our Example, $\left(i\right) : a = 25 = b \ne 0 , \left(i i\right) : h = 0 ,$ and,

$\left(i i i\right) : g = - 15 , f = 15 , c = - 18$

$\Rightarrow {g}^{2} + {f}^{2} - a c = 225 + 225 = 900 > 0$.

So, the eqn. represents a circle having centre #(15/25,-15/25), i.e.,

$\left(\frac{3}{5} , - \frac{3}{5}\right) \text{ & radius } \frac{\sqrt{900}}{|} 25 | = \frac{30}{25} = \frac{6}{5}$.

Enjoy Maths.!

Jan 14, 2017

Yes it is an equation of a circle.

#### Explanation:

Given: $25 {x}^{2} + 25 {y}^{2} - 30 x + 30 y - 18 = 0 \ldots \ldots \ldots \ldots \ldots E q n \left(1\right)$

If this an equation of a circle then we should be able to manipulate it back into standard form of ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
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I spot that the standard form is part of the process for completing the square. Lets investigate that.

Divide throughout by 25 to get rid of the coefficients for ${x}^{2} \mathmr{and} {y}^{2}$

${x}^{2} + {y}^{2} - \frac{6}{5} x + \frac{6}{5} y - \frac{18}{25} = 0$

Write as:

$\left({x}^{2} - \frac{6}{5} x\right) + \left({y}^{2} + \frac{6}{5} y\right) = + \frac{18}{25}$

Completing the squares

${\left(x - \frac{6}{10}\right)}^{2} - {\left(- \frac{6}{10}\right)}^{2} + {\left(y + \frac{6}{10}\right)}^{2} - {\left(+ \frac{6}{10}\right)}^{2} = \frac{18}{25}$

${\left(x - \frac{3}{5}\right)}^{2} + {\left(y + \frac{3}{5}\right)}^{2} = \frac{18}{25} + \frac{36}{100} + \frac{36}{100} = \frac{36}{25}$

Thus the centre as at point 1$\to {P}_{1} \to \left(x , y\right) = \left(\frac{3}{5} , - \frac{3}{5}\right)$

The radius is ${r}^{2} = \frac{36}{25} \implies r = \frac{6}{5}$

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$\textcolor{b l u e}{\text{Putting it all together}}$

$25 {x}^{2} + 25 {y}^{2} - 30 x + 30 y - 18 = 0$

Is the same as:

${\left(x - \frac{3}{5}\right)}^{2} + {\left(y + \frac{3}{5}\right)}^{2} = {r}^{2} = {\left(\frac{6}{5}\right)}^{2}$

Thus it is a circle. 