# Graphing Circle in Ti-83

#### Anemori

Can anybody help me graph me this equation in Ti-83 because it doesn't look right on me.

Here is the Equation:

y<=4-x^2 <---- parabola (I graph it ok)

16x^2+25y^2<=400

for the circle:

becomes:

x^2/25+y^2/16=1

$$\displaystyle y=+/- \frac{4\sqrt{-1(x-5)(x+5)}}{5(X-5)(x+5)}$$

is that right? then when input it in my TI-83 it doesn't look right...

pls help me out thanks!

#### skeeter

MHF Helper
Can anybody help me graph me this equation in Ti-83 because it doesn't look right on me.

Here is the Equation:

y<=4-x^2 <---- parabola (I graph it ok)

16x^2+25y^2<=400

for the circle:

becomes:

x^2/25+y^2/16=1

$$\displaystyle y=+/- \frac{4\sqrt{-1(x-5)(x+5)}}{5(X-5)(x+5)}$$

is that right? then when input it in my TI-83 it doesn't look right...

pls help me out thanks!
$$\displaystyle \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1$$ is an ellipse, not a circle.

equations to graph ...

$$\displaystyle y_1 = 4\sqrt{1 - \frac{x^2}{25}}$$

$$\displaystyle y_2 = -4\sqrt{1 - \frac{x^2}{25}}$$

• Anemori

#### Anemori

$$\displaystyle \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1$$ is an ellipse, not a circle.

equations to graph ...

$$\displaystyle y_1 = 4\sqrt{1 - \frac{x^2}{25}}$$

$$\displaystyle y_2 = -4\sqrt{1 - \frac{x^2}{25}}$$

Thanks!

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