Question

Determine the points of intersection of the equation circumference x² + (y-3) ² = 25 with the coordinate axes.

Answer 1

Answer:

$x^2 +(y-3)^2 = 25$

And we want to find the coordinate axes so then we can do the following:

If x=0 we have:

$0^2 +(y-3)^2 = 25$

$(y-3)^2= 25$

$y-3= \pm 5$

$y_1 = 5+3=8$

$y_2 = -5+3=-2$

Now of y =0 we have:

$x^2 +9 = 25$

$x^2 = 16$

$x= \pm 4$

And then the coordinate axes are:

(4,0) (-4,0), (0,8), (0,-2)

Step-by-step explanation:

For this cae we have the following functon given:

$x^2 +(y-3)^2 = 25$

And we want to find the coordinate axes so then we can do the following:

If x=0 we have:

$0^2 +(y-3)^2 = 25$

$(y-3)^2= 25$

$y-3= \pm 5$

$y_1 = 5+3=8$

$y_2 = -5+3=-2$

Now of y =0 we have:

$x^2 +9 = 25$

$x^2 = 16$

$x= \pm 4$

And then the coordinate axes are:

(4,0) (-4,0), (0,8), (0,-2)