How do you solve x^2+y^2=25 and y+5=1/2x^2  using substitution?

Apr 4, 2016

Solution set for $\left(x , y\right)$ is $\left(0 , - 5\right)$, $\left(4 , 3\right)$ and $\left(- 4 , 3\right)$

Explanation:

As $\frac{1}{2} {x}^{2} = y + 5$, we have ${x}^{2} = 2 \cdot \left(y + 5\right) = 2 y + 10$.

Putting this in ${x}^{2} + {y}^{2} = 25$, we get $2 y + 10 + {y}^{2} = 25$ or

${y}^{2} + 2 y - 15 = 0$ i.e. ${y}^{2} + 5 y - 3 y - 15 = 0$ or

$y \left(y + 5\right) - 3 \left(+ 5\right) = 0$ or $\left(y - 3\right) \left(y + 5\right) = 0$ i.e. $y = 3$ or $y = - 5$

Hence ${x}^{2} = 2 \cdot 3 + 10 = 16$ i.e. $x = \pm 4$ that is

${x}^{2} = 2 \cdot \left(- 5\right) + 10 = 0$ i.e. $x = 0$

Hence solution set for $\left(x , y\right)$ is $\left(0 , - 5\right)$, $\left(4 , 3\right)$ and $\left(- 4 , 3\right)$

Apr 4, 2016

$\textcolor{g r e e n}{\text{Points of intersection are:}}$

$\textcolor{b l u e}{\left(x , y\right) \to \left(- 4 , 3\right) \text{ ; "(+4,3)" ; } \left(0 , - 5\right)}$

Explanation:

Just an observation: ${x}^{2} + {y}^{2} = 25$ is the equation of a circle with the centre at the origin.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:
${x}^{2} + {y}^{2} = 25$ ...............................(1)
$y + 5 = \frac{1}{2} {x}^{2}$ ................................(2)

$\textcolor{b l u e}{\text{Solving for } y}$

Rewrite (1) as: ${x}^{2} = 25 - {y}^{2} \text{ } \ldots \ldots \ldots . . \left({1}_{a}\right)$
Rewrite (2) as: ${x}^{2} = 2 y + 10 \text{ } \ldots \ldots \ldots . . \left({2}_{a}\right)$

Equate $\left({1}_{a}\right) \text{ to "(2_a)" through } {x}^{2}$ This is the equivalent of substituting for ${x}^{2}$

$25 - {y}^{2} = 2 y + 10$

This is the same as

${y}^{2} + 2 y - 15 = 0$

$\left(y + 5\right) \left(y - 3\right) = 0$

" "color(blue)(=> y= -5" or " +3) .........................(3)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for } x}$

Substituting into equation $\left({2}_{a}\right)$. The y is not squared!

$\textcolor{b r o w n}{\text{condition 1 } y = - 5}$

${x}^{2} = 2 y + 10 \text{ " -> " } {x}^{2} = 2 \left(- 5\right) + 10$

" "color(green)(x^2=0" " =>" " x=0)
'...........................................
$\textcolor{b r o w n}{\text{condition 2 } y = + 3}$

${x}^{2} = 2 \left(3\right) + 10 = 16$

$\text{ } \textcolor{g r e e n}{x = \pm 4}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

The point of intersection are:

$\textcolor{b l u e}{\left(x , y\right) \to \left(- 4 , 3\right) \text{ ; "(+4,3)" ; } \left(0 , - 5\right)}$