Question

Find the points where the line y = − x 2 intersects a circle of radius 8 centered at the origin. Give exact values for the x and y coordinates.

Answer 1

Answer:

So the coordinates are $(1+ \sqrt{31}, 1-\sqrt{31}),(1-\sqrt{31} , 1+\sqrt{31})$

Step-by-step explanation:

For this case we assume that the line is given by this formula $y = 2-x$

We know that the general equation for a circle is given by:

$(x-h)^2 +(y-k)^2 = r^2$

Since the circle for this case is centered at the origin then h,k =0 and we have this:

$x^2 + y^2 = 8^2 = 64$   (1)

For this case we can replace the formula for y from the line into equation (1) and we got:

$x^2 + (2-x)^2= 64$

We know from algebra that $(a-b)^2 = a^2 -2ab +b^2$ if we use this concept we got:

$x^2 +4 -4x + x^2 = 64$

We can subtract 64 on both sides and we got:

$2x^2 -4x -60 =0$

Now we can divide both sides by 2 and we got:

$x^2 -2x -30 = 0$

And we can use the quadratic formula to solve this:

$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

For this case $a=1, b=-2 , c=-30$ and if we replace we got:

$x = \frac{2 \pm \sqrt{2-4*(1)*(-30)}}{2}$

And we got $x_1= 1+\sqrt{31} , x_2 = 1-\sqrt{31}$

Now we just need to replace into the original equation for the line and we get the y coordinates like this:

$y_1 = 2- (1+\sqrt{31}) =1-\sqrt{31}$

$y_2 = 2- (1-\sqrt{31}) =1+\sqrt{31}$

So the coordinates are $(1+ \sqrt{31}, 1-\sqrt{31}),(1-\sqrt{31} , 1+\sqrt{31})$